exercise 5.1: trigonometric ratios - kopykitab · exercise 5.1: trigonometric ratios 1.) find the...
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Exercise 5.1: Trigonometric Ratios
1.) Find the value of Trigonometric ratios in each of the following provided one of the six trigonometric ratios are given.
Sol.
(i)sinA=23sinA — |
Given:
sinA=23sin.A = | .....(1)
By definition,
SinA=23sin^4 = | = P e r p e n d i c u l a r H y p o t e n u s e ....(2)
By Comparing (1) and (2)
We get,
Perpendicular side = 2 and
Hypotenuse = 3
Therefore, by Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side
(AB)
Therefore,
32 = AB2 + 22
AB2 = 32 - 22
AB2 = 9 - 4
AB2 = 5
AB= V5\/5
Hence, Base = \5 \ /5
Now, COSA= BaseHypotenuseCOS A = Hyf™„use
COSA= \53 COS A =
Now, cosec A = isinA -r—r
Therefore,
Hypotenusecosec A = HypotenusePerependicular -=------ -77— 7—Perependicu lar
cosec A =321
HypotenuseNow, sec A = HypotenuseBase----Base-----
Therefore,
sec A = 3V5 -^=
Now, tan A = PerependicularBase PereP^^i ular
tan A = 2a/5
Now, cot A = BasePerpendicular -=-------T.— —Perpendicular
Therefore,
(ii) COsA—45cos A — |
Given: COsA=45cos^4. = .... (1)
By Definition,
COSA=BaseHypotenuseCOS A = „ Ba:se----- .... (2)Hypotenuse '
By comparing (1) and (2)
We get,
Base =4 and
Hypotenuse = 5
B A
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base (AB) and hypotenuse (AC) and get the perpendicular side
(BC)
52 = 42+ BC2
BC2 = 52 - 42
BC2= 2 5 -1 6
BC2 = 9
BC= 3
Hence, Perpendicular side = 3
Now,
■ * • « O , c / u c iS in A = 2 3 s m A = 3 = PerpendicularHypotenuse Hypotenuse
Perpendicular
Therefore,
sinA=35sinA = |
Now, cosec A = 1sinA -t—t-sinA
Therefore,
cosec A= 1 sinA —sinA
Therefore,
cosec A =HypotenusePerependicularHypotenuse
Perependicular
c o s e c A = 53
Now, sec A = 1 cosA — t’ cosA
Therefore,
sec A = HypotenuseBaseHypotenuse
Base
sec A = 54
. . . » _ „ PerpendicularNow, tan A = PerpendicularBase-----Base--------
Therefore,
tan A =34 44
Now, cot A = 1tanA — —7tanA
Therefore,
cot A = BasePerpendicular BasePerpendicular
COt A = 43 3
(iii) tan0=llltan© = y-
Given: ta n 0 = n ita n © = -y .... (1)
By definition,
. ^ ^ Perpendiculart a n 0 = PerpendicularBasetanH ------ .... (2)
By Comparing (1) and (2)
We get,
Base= 1 and
Perpendicular side= 5
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base side (AB) and perpendicular side (BC) and get hypotenuse(AC)
AC2 = 12 + 112
AC2 = 1 + 121
AC2= 122
AC= V l2 2 v /122
. , ■ PerpendicularNow, SIR ©— PerpendicularHypotenuse Sin 0 = 77— 7— —
Therefore,
sin0 =iiVi2 2 sin© = ^ =
Now, cosec0=lsinO@ —
cosec 0=V12211 0 =
Now, COS©= BaseHypotenuse cos 0 BaseHypotenuse
Therefore,
cosO= 1 \T22cos 0 = -^=
Now, secO=icososec 0 = —C O S 0
Therefore,
sec0= HypotenuseBasesec0 = sec©=V1221 sec 0 = sec©=Vl22
sec © = Vl22
Now, COtO=1tan0cot 0 =
Therefore,
COt0= BasePerpendicularCOt 0 — — — Ba']e , COtO=11lCOt© —Perpendicular l l
(iv) sin0=iH5sin© — ^
Given: s in © = lH 5 s in © = .... (1)
By definition,
Sin©— PerpendicularHypotenuse sin © PerpendicularHypotenuse . . . (2 )
By Comparing (1) and (2)
We get,
Perpendicular Side = 11 and
Hypotenuse= 15
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
152 = AB2 +112
AB2 = 152 - 112
AB2= 225-121
AB2 = 104
AB = Vl04\/l04
AB= V2x2x2x13\/2 x 2 x 2 x 13
AB= 2V2x13\/2 x 13
AB= 2V26\/26
Hence, Base = 2^26<\/26
Now, COS0= BaseHypotenuse COS © BaseHypotenuse
Therefore,
COS©= 2V2615 COS @ =
Now, cosec[©= ls in © 0 =s in 0
Therefore,
cosec©— HypotenusePerpendicular©Hypotenuse
Perpendicular
c o s e c © = 1 5 1 1 0 = - j j
HypotenuseNow, sec©— HypotenuseBase © — -----Base------
Therefore,
s e c © = 152V26 0 = ~ ^ =2y 26
Now, ta n © - PerpendicularBasetan 0 PerpendicularBase
Therefore,
tan©=ii2\26tan0 =
Now, COt0— BasePerpendicularCOt 0 = BasePerpendicular
Therefore,
COt©=2V2611 cot 0 = —qrj—■
(v)tana=5i2tana = ^
Given: tana=5l2ta im = ^ .... (1)
By definition,
tana— PerpendicularBase t a n OtPerpendicular
Base (2)
By comparing (1) and (2)
We get,
Base= 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base side (AB) and the perpendicular side (BC) and gte hypotenuse (AC)
AC2 = 122 + 52
AC2 = 144 + 25
AC2= 169
AC= 13
Hence Hypotenuse = 13
Now, Sina- PerpendicularHypotenusesin a = PerpendicularHypotenuse
Therefore,
s in a= 5 i3 s in a = ^
Now, COSecQ- HypotenusePerpendicularQ!Hypotenuse
Perpendicular
c o s e c a = i 3 5 a — -jr-
Now, COSQ— BaseHypotenuse COS Qt BaseHypotenuse
Therefore,
cosa=i2i3cosa — ^
Now, seca=icosaseca = —cosa
Therefore,
COtQ= BasePerpendicularcot O t = - = —Bas,e , C O ta = 1 2 5 c o ta = ^P e rv e n d ic u la r o
(vi) sin0=V 32sin© = ^
Given: sin©=V32sin@ = ^ .... (1)
By definition,
. . r\ Perpendicular / 0 .SI n 0 — P erpendicularHypotenuse S in 0 = Hypotenuse
By comparing (1) and (2)
We get,
Perpendicular Side = V 3 \/3
Hypotenuse = 2
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
22 = AB2 + (V3\/3)2
AB2 = 22 - (V3-s/3)2
AB2 = 4 - 3
AB2 = 1
AB= 1
Hence Base = 1
Now, COSO= BaseHypotenuse COS 0 = HypZnuse
Therefore,
COS©= 12COS 0 =
Now, COSecG=1sin0© — -r^ r
Therefore,
cosec©- HypotenusePerpendicualar© =Hypotenuse
Perpendicualar
Now, See©- HypotenuseBase sec 0Hypotenuse
Base
Therefore,
sec©=2 i sec© = j-
Now, ta n © - PerpendicularBasetan 0 = PerpendicularBase
Therefore,
tan0=V3i ta n © = ^
Now, COt0 — BasePerpendicularCOt 0 = BasePerpendicular
Therefore,
(vii) C O S © — 725COS © — ^T
Given: COS0=725cos@ = .... (1)
By definition,
COS©= BaseHypotenuse COS 0 „ B + StH y p o t e r
By comparing (1) and (2)
We get,
Base = 7 and
Hypotenuse = 25
Therefore
By Pythagoras theorem,
AC2= AB2 + BC2
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side(BC)
252 = 72 +BC2
BC2 = 252 - 72
BC2 = 625 - 49
BC = 576
BC= V576a/576
Hence, Perpendicular side = 24
Now, S iflO - perpendicularHypotenusesi.il 0 perpendicularHypotenuse
Therefore,
s in © = 242 5 s in 0 = ^
Now, cosecO=lsin©0 —sin 0
Therefore,
CO SecO - HypotenusePerpendicualar 0Hypotenuse
Perpendicualar
COSeC0= 2524 0 = | |
Now, secO=icos©sec0 = —^cos 0
Therefore,
S e C 0 = HypotenuseBasesec © — HVPotenus_e_ s e c @ = 2 5 7 sec 0 _
. , , ^ l PerpendicularNow, tan0- PerpendicularBasetan 0 = ---- ~Base--------
Therefore,
tan0=247ta n © — -y-
Now, COt0= 1tan0 cot 0 = - ^ 7 7tan 0
Therefore,
COt©= BasePerpendicularCOt 0 -=—Ba e~,— COt0= 724COt0Perpendicular
(viii) t a n 0 = 8 i 5 t a n 0 = ^
Given: tan©=8l5tan0 = .... (1)
By definition,
tanO- PerpendicularBasetan 0 = PerPe™dicul,ar (2 )
By comparing (1) and (2)
We get,
Base= 15 and
Perpendicular side = 8
Therefore,
By Pythagoras theorem,
AC2= 152 + 82
AC2= 225 + 64
AC2 = 289
AC = a/289^/289
AC= 17
Hence, Hypotenuse = 17
Now, S in0— PerpendicularHypotenusesin 0 PerpendicularHypotenuse
Therefore,
S in0=8l7sin@ = ^
Now, cosecO= ls in © 0 — -7 —-s in 0
Therefore,
COSecO— HypotenusePerpendicular 0Hypotenuse
Perpendicular
Now, COS0— BaseHypotenuse COS 0 = Hyp^Zuse
Therefore,
COSO= 1517 COS 0 — Yf
Now, Sec0=1cos0sec© = —cos 0
Therefore,
SeC0= HypotenuseBasesec0 — SeC0= 1715sec 0 —
Now, COtO= 1 tan© cot 0 = —’ tan©
Therefore,
C O t0= BasePerpendicularCOt 0 — —— Ba™ ,— C O tO =158C O t0Perpendicular
(ix) C O tO — 125 COt 0 = - y
Given: CO t0=i25cot© = ^ ■■■■ (1)
By definition,
COt©= 1 tan© cot 0 — — ^rtan ©
COt0= BasePerpendicularCOt 0 = -5— Ba*f ,— .... (2)
By comparing (1) and (2)
We get,
Base = 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
AC2= AB2 + BC2
Substituting the value of base side (AB) and perpendicular side(BC) and get the hypotenuse (AC)
AC2 = 122 + 52
AC2= 144 + 25
AC2 = 169
AC = V i 6 9 \/ l6 9
AC = 13
Hence, Hypotenuse = 13
Now, Sin©— PerpendicularHypotenusesin © PerpendicularHypotenuse
Therefore,
s in © = 5 i3 s in 0 = Tjj
Now, cosec©= ls in 0 @ =s in 0
Therefore,
COSecO— HypotenusePerpendicular©Hypotenuse
Perpendicular
cosec© = 1350 =
Now, C O S 0— BaseHypotenuse COS 0 BaseHypotenuse
Therefore,
COSO= 1213COS 0 — -jl
Now, secO= ic o s o s e c 0 =
Therefore,
SecO= HypotenuseBase sec 0 = SecO = 1312sec @ —
Now, tanO= 1 cot© ta n 0 = —cot ©
Therefore,
tanO=PerpendicularBasetan© = PerPe™dicular fa n © — 512ta n ©
(x) sec©=i35sec@ — ^
Given: sec0=l35sec@ = (1)
By definition,
sec©= i c o s o s e c © = — .... (2)
By comparing (1) and (2)
We get,
Base=5
Hypotenuse = 13
Therefore,
By Pythagoras theorem,
132 = 52 + BC2
BC2 = 132- 5 2
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side(BC)
BC2=169 - 25
BC2= 144
BC= V l4 4 \ / i4 4
BC = 12
Hence, Perpendicular side = 12
Now, Sin©— PerpendicularHypotenusesin 0 PerpendicularHypotenuse
Therefore,
s in © = i2 i3 s in © —
Now, cosec O = ls in © 0 =sin 0
Therefore,
C O S e c O — HypotenusePerpendicular©Hypotenuse
Perpendicular
c o s e c © = 1312 0 = j |
Now, COS0= 1 sec© cos 0 = —sec 0
Therefore,
COS0= BaseHypotenuse COS 0 = „ Bcf e---- COS0=513COS0 = ^Hypotenuse 13
Now, ta n 0 - PerpendicularBasetan 0 PerpendicularBase
Therefore,
tan 0 = 1 25 ta n © = 0
1tan ©
Now, COtO- 1tan0cot 0 =
Therefore,
COtO- BasePerpendicularCOt © — — — Base — COtO— 512 c o t 0 — -prPerpendicular IZ
Given:
(xi) cosecO=VlO© =
cosec0=Vioi0 = (1)
By definition
COSecO=1sin0@ = —r—77 ....(2)sin 0 ' '
0 — HypotenusePerpendicular© HypotenusePerpendicular
By comparing (1) and(2)
We get,
Perpendicular side= 1 and
Hypotenuse = V i OV^HT
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
(Vi 0 \/I0 )2 = a b 2 + 12
AB2= (V10x/1 0 )2 - 1 2
a b 2= 10 -1
AB = V9\/9
AB = 3
Hence, Base side = 3
Now, S iflO - PerpendicularHypotenuseSm ©Perpendicular
Hypotenuse
Therefore,
s in 0 = iV io s in © 1v/lo
Now, COS0— BaseHypotenuse COS © BaseHypotenuse
Therefore,
cos©= 3vTo cos 0 = - =■v/10
Now, sec©= icosOsec 0 = —cos 0
Therefore,
SeC0= HypotenuseBase sec 0 = SeC0=WlO3sec
. , . /-v ± PerpendicularNow, ta n 0 - PerpendicularBasetan 0 = ----------------
Therefore,
tan 0 = 13 ta n 0 =
Now, C O t O = Ita n O C O t 0 =
cot©= 31 c o t 0 = | cot0=3cot 0 = 3
( x i i ) C O S© 1215 c o s 0 ^ |
Given: C O S © 1 2 1 5 C O S 0 | | .... (1)
By definition,
COS0 — BaseHypotenuseCOS 0 = Hyp^ nuse ■■■■ (2 )
By comparing (1) and (2)
We get,
Base=12 and
Hypotenuse = 15
Therefore,
By Pythagoras theorem,
AC2 = AB2+ BC2
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
152 = 122 + BC2
BC2 = 152 - 122
BC2 = 225 - 144
BC 2= 81
bc = V81 V sT
BC = 9
Hence, Perpendicular side = 9
Now, SinO-PerpendicularHypotenusesin ©Perpendicular
Hypotenuse
Therefore,
sin0=9i5sin© _9_15
Therefore,
Now, cosec©= isin© © = -sin 0
cosec ©-HypotenusePerpendicular©Hypotenuse
Perpendicular
cosec©= 1 5 9 0 = y
Now, sec©=icososec© =
Therefore,
SeC0= HypotenuseBasesec0 = SeC0= 1512sec0 = |§
Now, ta n © - PerpendicularBasetan 0 = PerPe™dicular
Therefore,
tan©=9i2tan0 — y
Now, COt©= ItanOCOt 0 = —^ r’ tan ©
Therefore,
COt©= BasePerpendicularCOt © = —— Bas,e , C O t0=129C O t© = ^Perpendicular y
2.) In a AAABC, right angled at B , AB - 24 cm , BC= 7 cm , Determine
(i) sin A , cos A
(ii) sin C, cos C
Sol.
(i) The given triangle is below:
c
A
Given: In A A ABC , AB= 24 cm
BC = 7cm
Z A B C Z A B C = 90°
To find: sin A, cos A
In this problem, Hypotenuse side is unknown
Hence we first find hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
AC2 = AB2 + BC2
AC2 = 242 + 72
AC2 = 576 + 49
AC2= 625
AC = V625\/625
AC= 25
Hypotenuse = 25
By definition,
S in A — PerpendicularsideoppositetozAHypotenusesin A = Hypotenuse
sinA=BCACsinil = sinA=725sin A 7_25
By definition,
C O S A - BasesideadjacenttozAHypotenuse COS A =
c o s A = 41 C O S A = 2425 COS A — | |
BasesideadjacenttoZAHypotenuse c o s A = a bac
Answer:
sinA= 725sin A = cosA= 2425cos A = | |
(ii) The given triangle is below:
Given: In AA ABC , AB= 24 cm
BC = 7cm
zABCZABC = 90°
To find: sin C, cos C
In this problem, Hypotenuse side is unknown
Hence we first find hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
AC2 = AB2 + BC2
AC2 = 242 + 72
AC2 = 576 + 49
AC2= 625
AC = V625\/625
AC= 25
Hypotenuse = 25
By definition,
SinC— PerpendicularsideoppositetozCHypotenuse s in C —
sinC=a ba c s in C = sinC= 2425 s in C = U
Hypotenuse
By definition,
COSC— BasesideadjacenttozCHypotenuseCOS C
c o s A — COSA= 725 COS A — 4=r
BasesideadjacenttoZC COSA= BCACHypotenuse
Answer:
sinA= 2425sin A = C O SA =725C O sA = ^
3.) In the below figure, find tan P and cot R. Is tan P = cot R?
In the given right angled A P Q R A PQR, length of side OR is unknown
Therefore , by applying Pythagoras theorem in A P Q R A PQR
We get,
PR2 = PQ2 + QR2
Substituting the length of given side PR and PQ in the above equation
132= 122 + QR2
p
To find, tan P, cot R
Sol.
QR2 = 132 - 122
q r 2 = 169-144
QR2= 25
QR = V25v/25
By definiton, we know tha t,
tanP- PerpendicularsideoppositetozPBasesideadjacenttozP tan P
tanP=QRPQtan P =
PerpendicularsideoppositetoZPBasesideadjacenttoZP
ta n P = 5 i2 ta n P = .... (1)
Also, by definition, we know that
COtR— BasesideadjacenttozRPerpendicularsideoppositetozR
cot R — Base side adjacent to ZR Perpendicular side opposite to ZR
COtR=QRPQcot R QRPQ
co tR = 5 i2co tR = ^ .... (2)
Comparing equation (1) ad (2), we come to know that that R.H.S of both the equation are equal.
Therefore, L.H.S of both equations is also equal
tan P = cot R
Answer:
Yes , tan P =cot R = 512 ^
4.) If sin A = 941 Compute cos A and tan A.
Sol.
Given: s inA =94 ls inA = .... (1)
To find: cos A, tan A
By definition,
SinA— PerpendicularsideoppositetozAHypotenuse sin A Perpendicular side opposite to ZA Hypotenuse
By comparing (1) and (2)
We ge t,
Perpendicular side = 9 and
Hypotenuse = 41
Now using the perpendicular side and hypotenuse we can construct AABCAABC as shown below
Length of side AB is unknown is right angled AABCAABC ,
To find the length of side AB, we use Pythagoras theorem,
Therefore, by applying Pythagoras theorem in AABCAABC ,
We get,
AC2 = AB2 + BC2
412 = AB2 + 92
AB2 = 412 - 92
AB2 = 168-81
AB= 1600
AB = V1600^/1600
AB = 40
Hence, length of side AB= 40
Now
By definition,
COSA= BasesideadjacenttozAHypotenuse COS A
COS A = COSA= 4041 COS A =
Base side adjacent to ZA A------------rT— t---------------- COSA= ABACHypotenuse
Now,
By definition,
tanA— PerpendicularsideoppositetozABasesideadjacenttozA
ta n A = SECSl tanA=BCABtanA = -fg tanA=940tanA = £Base s*ae adjacent to Z A A B 4U
Answer:
C O S A = 4041 cos A = , tanA =940tan A — ^
5.) Given 15cot A=8, find sin A and sec A.
Answer:
Given: 15cot A = 8
To find: sin A, sec A
Since 15 cot A =8
By taking 15 on R.H.S
We get,
COtA=8i5cot A = ^
By definition,
COtA= ItanACOt A — -^—r
Hence,
cotA= 1 PerpendicularsideoppositetozABasesideadjacenttozA cot A __________1_________Perpendicular side opposite to ZA
Base side adjacent to ZA
COtA— BasesideadjacenttozAPerpendicularsideoppositetozA
Base side adjacent to Z Acot A —
Perpendicular side opposite to Z A
Comparing equation (1) and (2)
We get,
Base side adjacent to Z A Z A = 8
Perpendicular side opposite to Z A Z A = 15
A A B C A A B C can be drawn below using above information
Hypotenuse side is unknown.
Therefore, we find side AC of AABCAABC by Pythagoras theorem.
So, by applying Pythagoras theorem to AABCA A B C
We get,
AC2 = AB2 +BC2
Substituting values of sides from the above figure
AC2 = 82 + 152
AC2 = 64 + 225
AC2 = 289
AC = V289>/289
AC = 17
Therefore, hypotenuse =17
Now by definition,
S in A — PerpendicularsideoppositetozAHypotenuse s in APerpendicular side opposite to / .A
Hypotenuse
Therefore, sinA=BCACsin A — ^’ A C
Substituting values of sides from the above figure
sinA=i5i7sinA =
By definition,
secA=icosAsec = -A _cos A
Hence,
secA= 1 BasesideadjacenttozAHypotenuse S e C A ________ 1________Base side adjacent to Z.A
Hypotenuse
SecA— HypotenuseBasesideadjacenttozA se c A — HypotenuseBase side adjacent to / .A
Substituting values of sides from the above figure
secA=i78sec A = ^
Answer:
sinA=i5i7sinz4 — secA=i78secz4 — ^
6.) In AP QRAPQR , right angled at Q, PQ = 4cm and RQ= 3 cm .Find the valueof sin P, sin R , sec P and sec R.
Sol.
Given:
A P Q R A PQR is right angled at vertex Q.
PQ= 4cm
RQ= 3cm
To find,
sin P, sin R , sec P , sec R
Given APQRAPQR is as shown below
p
<1
Hypotenuse side PR is unknown.
Therefore, we find side PR of APQRA PQR by Pythagoras theorem
By applying Pythagoras theorem to APQRA PQR
We get,
PR2 = PQ2 +RQ2
Substituting values of sides from the above figure
PR2 = 42 +32
PR2 = 25
PR = V25V25
PR = 5
Hence, Hypotenuse =5
Now by definition,
Substituting values of sides from the above figure
sinP=35sinP — |
PR2 = 16 + 9
SinP — PerpendicularsideoppositetozPHypotenuse s in P =
SinP = RQPRsin P =
Hypotenuse
Now by definition,
S in R — PerpendicularsideoppositetozRHypotenuse s in RPerpendicular side opposite to Z R
Hypotenuse
sinR=PQPRsin# —
Substituting the values of sides from above figure
sin R= 45 sin# = 40By definition,
secP = icosP se c P = ----------- S e C P — 1 BasesideadiacenttozpHypotenuse SeC P — —5------ r;— y . ------ 7 Z — /—1 r 1 r Base side adjacent to ZipHypotenuse
S e c P — HypotenuseBasesideadjacenttozP se c P — HypotenuseBase side adjacent to Z P
Substituting values of sides from the above figure
secP = prpqsec P = ^ secP= 5 4 sec P = f
By definition,
secR=icosRsec# 1cos R
secR= 1BasesideadjacenttozRHypotenuse S e C R ______1______Base side adjacent to ZR
Hypotenuse
SecR— HypotenuseBasesideadjacenttozRsec R HypotenuseBase side adjacent to Z R
Substituting values of sides from the above figure
secR=PRRQsec# — secR =53sec# = |
Answer:
sinP =35sinP = | , sinR=45sin# =
secP=54sec P = secR=53sec# = |
7.) If cot0=78cot @ = | , evaluate
(i) 1+sin0xl-s in0 1+cos0x1-cos0 1+sin Q x 1-sin 0 1+cos 0 x 1-cos 0
(ii) COt20cot2 0
Sol.
Given: C O t© = 7 8 c o t© —
To evaluate: l+sin0xl-sin0l+cos0xl-cos0 1+ sin® x sin®l+ co s 0 x l-cos 0
. . _ . . _ . _ 1+sin 0 x 1-sin 0 ,l+ co s 0 x 1-cos 0 ' '
We know the following formula
(a + b)(a - b) = a2 - b2
By applying the above formula in the numerator of equation (1)
We get,
(1 +sin0)x(1-sin0)=1-sin20.... (2)(Where,a=1 andb=sin0)(1 + sinO) x (1 -sinff) = 1 -sin29........ (2) (Where, a = 1 and b = sin0)
Similarly,
By applying formula (a +b) (a - b) = a2 - b2 in the denominator of equation (1).
We get,
(1+cos0)(1— cos0)=12—cos20 ( l + cos@ )(l-cos@ ) = l 2-cos2 0 ... (Where a=1
and b= COS0cos 0
(1+COS0)(1—COS0)=1—COS20 (1 + c o s 0 )( l-c o s 0 ) — 1-cos2 0 ... (Where a=1 and
b= cosOcos 0
Substituting the value of numerator and denominator of equation (1) from equation (2), equation (3).
Therefore,
(1 +sin0)(1-sin0)(1 +cos0)(1-cos0) (1+sm^ ^ ~ Sm^ = 1-sin201-cos20 1-sin ® ....(4 )' A A A ' (l+ cos 0 ) (1-cos 0 ) l-co s2 0 v ’
Since,
cos20+sin20=1cos2 © + sin2 0 = 1
Therefore,
cos20=1 -s in2 0cos2 0 = 1-sin2 0
Also, sin20 = 1 -cos20sin2 0 — 1-cos2 0
Putting the value of 1— sin2© 1 - sin2 © and 1— COS201-co s2 © in equation (4)
We get,
(l+sin 0 )(l-sin 0 ) „ , cos2 0(1+sin0 )(1-sin0 )(1+cos0 )(1-cos0 ) 77-------777-:-------= cos20 sin20 — 5—(1+COS0) (1-cos 0 ) sin2 0
We know that, cos0sin0=COt©-er L^- — cot©sin 0
(1 +sin0)(1-sin0)(1 +cos0)(1-cos0) q ) = ( C o t 0 )2 ( c o t 0 ) 2
Since, it is given that COt©=78cot 0 = |
Therefore,
(1 +sin0)(1-sin0)(1 +cos0)(1-cos0)(l+sin 0 ) (1-sin 0 ) (1+cos 0 ) (1-cos 0 ) = (78)2( | ) 2
(1 +sin0)(1-sin0)(1 +cos0)(1-cos0)(l+sin 0 ) (1-sin 0 ) _ 2. 2 ^(1+cos 0 ) (1-cos 0 ) 82
(1 +sin0)(1-sin0)(1 +cos0)(1-cos0)(l+ sin 0 ) (1-sin 0 )
(1+cos 0 ) (1-cos 0 ) = 49641
(ii) Given: COt©=78cot 0 = |
To evaluate: COt20co t2 ©
COt0= 78 COt © — |
Squaring on both sides,
We get,
(cot©)2=(78)2(c o t0 )2 = ( | ) 2
(COt©)2(cot 0 ) 2 =4964 | |
Answer:
4949641
1—tan2 A _ COg2 s jn 2 ^ Qr nQj 1+tan2 A
Sol.
Given: 3cot A =4
To check whether i-tan2Ai +tan2A=COS2A—sin2A 1~tan*\At — cos2 A -s in 2 A or not.l+ ta n 2 A
8.) If 3cotA=43 cot A — 4 , check whether i-tan2Ai+tan2A=COS2A—sin2A
3cot A =4
Dividing by 3 on both sides,
We get,
cot A = 431 .... (1)
By definition,
COtA= ItanACOt A — —tan A
Therefore,
COtA= 1 PerpendicularsideoppositetozABasesideadjacenttozA COt A = Perpendicular opposite to ZA~Base side adjacent to Z A
CO tA— BasesideadjacenttozAPerpendicularsideoppositetozA
Base side adjacent to Z Acot A —Perpendicular side opposite to / .A -■ (2 )
Comparing (1) and (2)
We get,
Base side adjacent to Z A Z A = 4
Perpendicular side opposite to z A Z A = 3
Hence AABCAABC is as shown in figure below
In AABCAABC , Hypotenuse is unknown
Hence, it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem inAABCAABC
We get
AC2= AB2 +BC2
Substituting the values of sides from the above figure
AC2 =42 + 32
AC2 = 16 +9
AC2 = 25
AC = V25v/25
AC = 5
Hence, hypotenuse= 5
To check whether
To check whether l-tan2Al +tan2A=COS2A—sin2A }~tWL\A. = cos2 A -s in 2 A or not.l+ tan2 A
We get thee values of tan A , cos A , sin A
By definition,
tan A = icotA—cot A
Substituting the value of cot A from equation (1)
We get,
tan A = 14 -j4
tan A = 3 4 4 .... (3)
Now by definition,
* a Base side adjacent to ZA ACOSA— BasesideadjacenttozAHypotenuse COS A = --------TT ^___ ____ COSA— ABACHypotenuse
COS A = # AC
Substituting the values of sides from the above figure
COSA=45COS A = | .... (5)
Now we first take L.H.S of equation l-tan2Al+tan2A=COS2A—sin2A
l W A = c o s 2 ^ _ sip2 ^1+tan2 A
L.H.S = 1-tan2A1+tan2A 1-tan2 A 1+tan2 A
Substituting value of tan A from equation (3)
We get,
L.H.S= 1- ( 34)21+(34)2H f ) 2i+ ( l) 2
l - f - l21-tan2A1 +tan2A 1 tan2^ = 1—(34)21 +(34 )2 -----
1+tan2 A v ' v ' l + ( - ) 2
j __ 9_
1-tan2A1+tan2A } tan,^ = 1—9161+916 161+tan2 A 1 + —^ 16
Taking L.C.M on both numerator and denominator
We get,
1-tan2A1 +tan2A } tan2„^ 1+tan2 A
= 16-91616+916
16-916
16+916
725 ■■■■ (8)
Now we take R.H.S of equation whether 1-tan2A1+tan2A=COS2A—sin2A
= cos2 A -s in 2 A1+tan2 A
R.H.S = cos2A-sin2Acos2 A - sin2 A
Substituting value of sin A and cos A from equation (4) and (5)
We get,
R . H . S = ( 4 5 ) 2- ( 3 5 2 ) ( f ) 2 - ( f 2 )
cos2A-sin2Acos2 A - sin2 A= (45)2-(352) ( | ) 2- ( | 2)
cos2A-sin2Acos2 A - sin2 A = 16 25- 9 2 5 1 |-
cos2A-sin2Acos2 A - sin2 A = 16 -9 25
cos2A-sin2Acos2 A-sin2 A = 7 2 5 ^ ....(7)
Comparing (6) and (7)
We get.
1-tan2A1 +tan2A=cos2A-sin2A }~tanl AA = cos2 A-sin2 Al+ ta n 2 A
Answer:
Yes, 1 -tan2Ai +tan2A =cos2 A—s i n2 A A. — cos2 A-sin2 Al+ ta n 2 A
9.) If tanO =abtan@ = f0 find the value of cosO+sinOcosO-sinO cos Q+sin Q cos 0 -s in 0
Sol.
Given:
tan© =abtan© — | .... (1)
Now, we know that tan©=sin0cos0tan ©
Therefore equation (1) become as follows
. _ _ sin© _ . asinOcos©----- 77 - ab -rcos 0 0
Now, by applying invertendo
We get,
. cos 0 b cosOsinO — ba . „ — — sin 0 a
Now by applying Componendo - dividendo
We get,
cos0+sin0cos0-sin0 — b+ab-acos 0 + s in 0 cos 0 -s in 0
b+ab-a
Therefore,
cosO+sinOcosO-sinO = b+ab-a cos 0 + s in 0 cos 0 -s in 0
b+ab-a
10.) If 3tanO=43 tan 0 = 4 , find the value of 4cosO-sin02cosO+sinO
Sol.
Given: If 3tan©=43 tan 0 = 4
Therefore,
tan© =43tan© = | .... (1)
Now, we know that tan©=sin0cos0tan 0 —cos 0
Therefore equation (1) becomes
sinQcosQ = 43 sin® 4 ■■■■(2)cos 0 3 ' '
Now, by applying Invertendo to equation (2)
We get,
cosQsinQ = 34 cos® = 4 .... (3) sin 0 4 v '
Now, multiplying by 4 on both sides
We get
4x cosOsinO =4X34 4 X ^ 7 7 = 4 X 7sin 0 4
Therefore
4cos0-sin0sin0 = 3-11 4 0 -s in 0 = 3=1sm 0 1
4 cos 0 -s in 0 2 cos 0 + s in 0
4cos©-sin©sin© = 21 4cos.e ^ in 0 = | .... (4)sin © 1 v '
Now, multiplying by 2 on both sides of equation (3)
We get,
2cos0sin0 = 32 2cos0 = f sin 0 2
Now by applying componendo in above equation
o 0 0 2cos 0 + s i n 0 3+22cos0+sin0sin0 - 3 + 2 2 -------- :——--------- = —F—sin 0 2
„ ^ ^ ^ 2 cos 0 + s in 0 52cos0+sin0sin0 — 5 2 -------t ~ ^ ------- = — ....(5 )sin 0 2 ' '
We get,
4cos©-sin0sin0 2cos0+sin0sin0 — 21 52
4 c o s 0 - s i n 0
s in 6
2 c o s 0 + s i n 0
s in 6
21_5
2
Therefore,
4cos0-sin0sin0 x sin02cos0+sin0 = 21 x 25 4 cos e ^ m 0sin 0
X sin 02 cos 0 + s in 0
21
Therefore, on L.H.S s inO sin© cancels and we get,
4cos0-sin02cos0+sin0 = 21 x 25 4 cos0~sin 0 = 7 X f2 cos 0 + s m 0 1 5
Therefore,
4cosO—sinO=44 c o s 0 - s in 0 = 4
11.) If 3cotO=23 cot 0 = 2, find the value of 4sinO-3cosQ2sinO+6cosO
Sol.
Given:
3cotO=23 cot 0 = 2
Therefore,
C O t O = 2 3 c o t 0 = | . . . . (1 )
25
4 sin 0 - 3 cos 0 2 sin 0 + 6 cos 0
Now, we know that COt0=cos0sin0cot © — sin ©
Therefore equation (1) becomes
cos0sin0 = 23 c?s® ....(2 )sin © o v 7
Now , by applying invertendo to equation (2)
sin0cos0 = 3 2 | ....(3)cos © 2, v 7
Now, multiplying by 43 4 on both sides,O
We get,
43 x sin0cos0 = 43 x 32 4 X = 4 X |3 cos 0 3 2
Therefore, 3 cancels out on R.H.S and
We get,
. . _.0 4 sin© 24sin03cos0 — 21 ------ rr T3 cos 0 1
Now by applying invertendo dividendo in above equation
We get,
. . _ 0 „ . . 4 sin 0 - 3 cos 0 2-14sin0-3cos03cos0 — 2 -1 1 ----- ------ =------ — —3 cos © 1
„ . _ 0 _ _ AA 4 sin 0-3 cos 0 1 tA\4sin0-3cos03cos0 — 11----- ~ ^ ------ = -r . . . .(4)3 c o s 0 1 ' '
Now, multiplying by 261 on both sides of equation (3)
We get,
26 x sin0cos0 = 26 x 32 -§ X = | X |6 cos 0 6 2
Therefore, 2 cancels out on R.H.S and
We get,
„ . ______ 2 sin 0 3 „ . ^ „ 2 sin 0 12sin06cos0 — 36 ----- tt -t 2sin06cos0 — 12 ----- -t6 cos 0 b 6 cos 0 2
Now by applying componendo in above equation
We get,
2cos0+6sin06sin0 — 1 +222 cos 0 + 6 sin 0
6 sin 01+2
2
2cos0+6sin06sin0 — 32 2 cos 0 + 6 sin 0 6 sin 0
32 ' ...(5)
Now, by dividing equation (4) by (5)
We get,
4 sin 0 - 3 c o s 0 1
_ 3 s in 0 ____ 14sin0-3cos03sin02cos0+6sin06sin0 — 11 32 - r ------~ c .—pr- — ~~T
2 c o s 0 + 6 s in 0 £
6 s in 0 2
Therefore,
4sin0-3cos03sin0 * 6sin02cos0+6sin0 = 11 * 23 4 sin e 73 c°s 0 x - — 6-fin 0 . _ = \3 sin 0 2 cos 0 + 6 sin 0 1
4sin0-3cos03sin0 * 2x3sin02cos0+6sin0 = 11 * 23 4 sm® 73 ™ s 0 X 2x^ ^ 0 =3 sin 0 2 cos 0 + 6 sin 0
Therefore, on L.H.S (3 Sin0sin0) cancels out and we get,
„ , . ^ ^ ^ „ v „ „ 2 x 4 sin 0 - 3 cos 0 1 w 22x4sin0-3cos02cos0+6sin0 — 11 x 23 —-----_ , _ . _ = 7 X 72 cos 0 + 6 sm 0 1 3
Now, by taking 2 in the numerator of L.H.S on the R.H.S
We get,
. . 0 „ 4 sin 0 -3 cos 0 24sin©-3cos02cos0+6sin0 — 23x2 -------_ , _ . _ = ^ —772 c o s 0 + 6 s m 0 3x2
Therefore, 2 cancels out on R.H.S and
We get,
. . 0 _ . 0 4 sin 0 - 3 cos 0 14sin©-3cos02cos0+6sin0 —13 7;------„ , „ . _ ■=■2 cos 0 + 6 sm 0 3
Hence answer,
. . _ . _ . _ . _ _ . _ 4 sin 0 - 3 cos 0 14sinO-3cos02cosO+6sinO —13 ------ „ , _ . _ — -5-2 cos 0 + 6 sin 0 3
12.) If tanO = abtan0 = ^ , prove that asin0-bcos0 asin0 +bcos0 = a2-b2a2+b2
a sin 0 - 6 cos 0 __ a2-6 2a sin 0 + 6 cos 0 a2+ 6 2
Sol.
Given:
ta n O a b ta n © ! .... (1)
Now, we know that tan©=sin0cos0tan © =cos 0
Therefore equation (1) becomes
sin0cos0 = ab sin® ^ ■■■■(2)COS 0 0 v 7
Now, by multiplying by ab | on both sides of equation (2)
We get,
abx sin0cos0 = ab Xabf- X ^ 7 7 = x f-6 cos 0 o o
Therefore,
asin0bcos0 — a2b2 ° sin® ....(3)6 cos 0 b2
Now by applying dividendo in above equation (3)
We get,
asin0-beos0bcos0 — a2-b2b2 asm'9 b5.osQ = a ■■■■(4)o cos© b2.
Now by applying componendo in equation (3)
We get,
asin0+bcos0bcos0 — a2+b2b2a sin 0 + 6 cos 0 6 cos 0
a2+6262
....(5)
Now, by dividing equation (4) by equation (5)
We get,
asin©-bcos0 bcos0 asin0 +bcos0 bcos0 — a2-b 2b2 a2+b2b2
a s in 0 - b c o s Q
b c o s Q
a s in Q + b c o s Q
b c o s ©
c?-&ft2
a 2 + 6 2
ft2
Therefore,
asin0-bcos0bcos0 x bcos0asin0+bcos0 = a2-b2b2 x b2a2+b2
a sin 0 - 6 cos 0 v 6 cos 0 __ a2—b2 626 cos 0 a sin 0 + 6 cos 0 62 a2 + 6 2
Therefore, bcosObcos © and b2 cancels on L.H.S and R.H.S respectively
asin©-bcos0asin0+bcos0 — a2-b 2a2+b2 a sin 0 -6 cos Q a sin 0 + 6 cos 0
a2—l
a2+62
Hence, it is proved that
asin0 -bcos0 asin0 +bcos0 = a2-b 2a2+b2 a sin 0 -6 cos 0 a sin 0 + 6 cos 0
a2—62 g2+62
13.) If sec0=135sec@ = show that 2sinQ-3cosQ4sinQ-9cosQ =3 l cos ®7 5 4 sin 0 - 9 cos 0
Sol.
Given:
sec0=i35sec@ — 5
To show that 2sin0-3cos04sin0-9cos0 =3 \ s|n ® j? cos ® = 34 sin 0 - 9 cos 0
Now, we know that COS©= lsec0cos © — — ^rsec 0
Therefore,
COS©= 1135 COS 0 = -jj-T
Therefore,
COS0=513c o s@ = ^ .... (1)
Now, we know that
C O S 0 — Basesideadjacenttoz0Hypotenuse COS 0Base side adjacent to ZO
Hypotenuse
Now, by comparing equation (1) and(2)
We get,
Base side adjacent to Z 0 Z © = 5
And
Hypotenuse =13
Therefore from above figure
Base side BC = 5
Hypotenuse AC = 13
Side AB is unknown. It can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2 = AB2 + BC2
Therefore by substituting the values of known sides
We get,
132 = AB2 + 52
Therefore,
AB2= 132 - 52
AB2= 169-25
AB2 = 144
AB = V144VT44
Therefore,
AB = 12 .... (3)
Now, we know that
s in© = abac s in © =A C
s in © = i2 i3 s in © = .... (4)
Now L.H.S of the equation to be proved is as follows
L.H.S = 2sinQ-3tanQ4sinQ-3cosQ \ sln ® 1*tan ®4 sin 0 - 3 cos 0
Substituting the value COS©cos 0 of sinOsin 0and from equation (1) and (4) respectively
We get,
2x 1213— 3X5134* 1213— 9*5132x — -3 x — 13 6 x 13
4x — -9 x —13 y X 13
Therefore,
L.H.S = 2 x 1 2 - 3 x 6 4 x 1 2 - 9 x 5
L.H.S = 24-1548-45
L.H.S= 931
L.H.S= 3
Hence proved that,
_ . _ _ . . _ _ _ 2 sin0 - 3 ta n © _ 02sinO-3tan04sinO-3cosO . . _ „---- rr- = 34 sin 0 - 3 cos 0
14.) If c o s 0 = 1 2 1 3 c o s 0 = , show that sinO(1-tanO)=35i56
sin 0 ( 1 - ta n 0 ) — ^
Sol.
Given: COS0=1213COS0 = | | .... (1)
To show that S in0(1—tan0)=35l56sin 0 (1 —ta n 0 ) =
Now we know that COS0= BasesideadjacenttozGHypotenuse COS 0 =
....(2)
Therefore, by comparing equation (1) and (2)
We get,
Base side adjacent to Z 0 Hypotenuse
Base side adjacent to Z 0 Z 0 = 12
And
Hypotenuse = 13
Therefore from above figure
Base side BC= 12
Hypotenuse AC= 13
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2= AB2 + BC2
Therefore by substituting the values of known sides
We get,
132= AB2 + 122
Therefore,
AB2= 132- 122
AB2= 169-144
AB =25
AB= V25\/25
Therefore,
AB = 5 .... (3)
Now, we know that
................. . Perpendicular side opposite to Z 0SinW- PerpendicularsideoppositetozOHypotenuseSin fc) = ---------- jj—~i------------------------
Now from figure (a)
sin©=abac s in © —A C
Therefore,
S in© =5 l2s in0 = ^ .... (5)
Now L.H.S of the equation to be proved is as follows
L.H.S of the equation to be proved is as follows
L.H.S = s in0(1-tan0]s in@ (l-tan@ ] .... (6)
Substituting the value of sinOsin 0 and tanOtan 0 from equation (4) and (5)
We get,
L.H.S = 513(1-512)^(1 -
Taking L.C.M inside the bracket
We get,
L-H.S= 513(lx121«12 -512 )^(ifif -
Therefore,
L.H.S= 513(12-512)1| ( ^ - )
L.H.S= 5 1 3 (7 1 2 )^ (^ )
Now by opening the bracket and simplifying
We get,
L.H.S= 5 x 7 1 3 x 1 2 ^
L.H.S= 35136^
From equation (6) and (7) ,it can be shown that
that sin0(1—tan0)sin 0 ( 1 - ta n 0 ) = 35136^
15.) If C O t© =lV 3C O t 0 = -4= , show that 1-cos202-sin20 = 35 1~cos„ ® f’ 2-sin2 0 5
We get,
Sol.
Given: COt©= W3cot @ ..(1)
To show that 1-cos202-sin20 = 35 1 cos2 0 2-sin2 0
35
N o w , w e k n o w that COtO= 1 t a n 0 c o t 0 - —tan 0
Since ta n © - Perpendicularsideoppositetoz0Basesideadjacenttoz0
tan© = Perpendicular side opposite to ZQ Base side adjacent to ZQ
Therefore,
ta n © = i PerpendicularsideoppositetozOBasesideadjacenttozQ tan© ________ l________Perpendicular side opposite to Z Q
Base side adjacent to ZQ
Therefore,
C O t0—Basesideadjacenttoz.0Perpendicularsideoppositetozi0
Base side adjacent to ZQCOt O Perpendicular side opposite to ZQ
Comparing Equation (1) and (2)
We get.
Base side adjacent to Z .0 Z 0 = 1
Perpendicular side opposite to Z.QZQ =
Therefore, triangle representing angle V 3\/3 is as shown below
Therefore, by substituting the values of known sides
We get,
AC2 = (V3)2(-s/3 )2 + 12
Therefore,
AC2 = 3+1
AC2= 4
AC= V 4 \/4
Therefore,
AC = 2 .... (3)
Now, we know that
„■ „ ................. . Perpendicular side opposite to Z 0S in W - PerpendicularsideoppositetozOHypotenuseSin fc) = -------------- jj—~i------------------------r Hypotenuse
Now from figure (a)
S i n 0 = A B A C s i n © = AC
Therefore from figure (a) and equation (3),
s in©=\32sin 0 = ^
Now we know that
C O S 0 BasesideadjacenttozOHypotenuse COS 0Base side adjacent to Z 0
Hypotenuse
Now from figure (a)
We get,
BCBCAC AC
Therefore from figure (a) and equation (3),
C O S 0 = 12COS 0 = \ . . . . (5)
Now, L.H.S of the equation to be proved is as follows
L.H.S = 1-cos2Q2-sin2Q 1 cos„ 02-sin2 0
Substituting the value of from equation (4) and (5)
We get,
Now by taking L.C.M in numerator as well as denominator
We get,
( 4 x l ) - l
L.H.S= (4x1H 4(4x2)-34
4
Therefore,
4-1L.H.S = 4-148-34 -^g-
4
L.H.S = 34x 45 X |
L.H.S = 35 4 = R.H.S0
Therefore,
A ■ 2 n - o r 1-COS2 0 31-cosz02-sinz0 — 3 5 --------5— -=-2 -s in 2 0 5
16.) lftan0=iV7tan0 = then show that cosec20 -sec20 cosec20 +sec20
cosec2 0 -se c 2 0 _ ^ jl cosec2 0 + sec2 0 4
Sol.
Given: tan©=lV7tan@ = ~ = .... (1)
To show that cosec20 -sec2Ocosec20+sec20 cosec ® s e c ® = 34 4cosed10+sec^ 0 4
Now, we know that
Since, tan©— PerpendicularsideopositetozOBasesideadjacenttozO
tail 0 _ Perpendicular side oposite to Z 0Base side adjacent to Z 0
Therefore,
Comparing equation (1) and (2)
We get.
Perpendicular side opposite to Z0Z@ = 1
Base side adjacent to Z0Z@ = V7y/7
Therefore, Triangle representing Z0Z@ is shown below
Hypotenuse AC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2= AB2 + BC2
Therefore by substituting the values of known sides
We get,
AC2= (1)2 + (V7)2( \ /7 )2
Therefore,
AC 2 = 1 +7
AC2 = 8
AC = y l S V 8
AC = V2x2x2V2 x 2 x 2
Therefore,
AC = 2V22\/2 .... (3)
Now we know that
SinO-PerpendicularsideoppositetozOHypotenuse s in 0
sin0=a ba c s in 0 —
Perpendicular side opposite to Z 0 Hypotenuse
sinO = i2V 2sin0 = .... (4)
Now, we know that cosec 0 = isin© 0 lsin 0
Therefore, from equation (4)
We get,
cosec 0 = 2 ^ 2 0 = 2y/2 .... (5)
Now, we know that
COS©= BasesideadjacenttozOHypotenuse COS 0Base side adjacent to Z 0
Hypotenuse
Now from figure (a)
We get,
CO S©=BCACCO S© = ^ 7
Therefore from figure (a) and equation (3)
C O S© =V72V2CO S0 = .... (6 )
Now we know that sec0= 1 cos© sec 0 = —cos 0
Therefore, from equation (6)
We get,
sec©= 1V72V2 sec 0 =2s/2
sec©= 2\2\7s e c 0 — .... (7)
Now, L.H.S of the equation to be proved is as follows
L.H.S cosec20 -sec20cosec20+sec20 cosec2 0 -se c 2 0 cosec2 Q + sec2 0
Substituting the value of cosec0© andsecOsec 0 from equation (6) and (7)
We get,
2 2 — 2 2 [(V2)]1‘-{^4)'L.H.S = [(2V2)] -(2V2V7) [(2V2)] +(2V5V7) v
[(2V2)]S+ ( 2 f )
L.H.S= (8 ) _ ( 87 ) (8 )+ ( 87 )w - d )
(8>+d)
Therefore,
56-8756+87
5 6 -8
7
5 6 + 8
48
L.H.S = 487647 -g j-
Therefore,
L.H.S = 4864 o4
L.H.S = 34 4 = R.H.S
Therefore,
cosec20 -sec20cosec20+sec20 cosec2 0-sec2 0 _ cosec2 0+sec2 0
Hence proved that
cosec20-sec20cosec20+sec20 cosec0^ ~ sec0f .cosec2 B+sec2 B
17.) If sec0=54sec 0 = |,find the value of sin0-2cos0tan0-cot0
Sol.
Given: sec0=54sec0 = -| .... (1)
To find the value of sin0-2cos0tan0-cot0 sm ® 2 cos ®tan 0 -c o t 0
Now we know that sec©= icos0sec 0 cos 0
sin 0 -2 cos 0 tan 0 -cot 0
Therefore,
C O S O = 1sec0 COS 0 —
Therefore from equation (1)
COSO= 15COS 0 =
C O S0=45C O S@ — | .... (2)
Also, we know that COS2O+sin20=1cos2 0 + sin2 0 = 1
Therefore,
sin20 = 1 -cos2Osin2 0 = 1-cos2 0 s in0=V l—cos2Osin© — V I-c o s 2 0
Substituting the value of COS0cos ©from equation (2)
We get,
Therefore,
s inO =35sin0 = | .... (3)
Also, we know that
sec20=1 +tan20sec2 0 = 1 + ta n 2 0
Therefore,
Therefore,
ta n 0 = 3 4 ta n 0 = | .... (4)
Also, COt©= ItanOcot 0 1tan 0
Therefore from equation (4)
We get,
COt©=43COt© = | .... (5)
Substituting the value of COS0cos 0 , COtOcot 0 and tanOtan© from the equation (2),(3),(4) and (5) respectively in the expression below
sin0-2cos0tan0-cot0 sin 0 -2 cos 0 tan 0 -cot 0
We get,
I — 2(±)sin0-2cos0tan0-cot0 S*D ® 2 cos ® = 35- 2 (45)3 4 -4 3 —j —tan 0 -cot 0 ' ' £ _ i
Therefore, sinQ-2cosQtanQ-cotQ sn ® 2 cos ® = 127 4tan 0 -cot 0 7
18.) If S in 0 = l2 l3 s in 0 = , find the value of 2sinQcosQcos2Q-sin2Q 2 slon ® cos®cos2 0 -s in ©
Sol.
Given: S in 0 = l2 l3 s in 0 = j | .... (1)
To, find the value of 2sin0cos0cos20-sin20 2 si„n ® cos ®cos2 0-sin2 0
Now, we know the following trigonometric identity
cosec2 0=1 +tan2© 0 = 1 + ta n 2 0
Therefore, by substituting the value of tanO tan 0 from equation (1)
We get,
2 2cosec2 0 = 1 + ( 12 13 ) 0 = 1 + ( ^ | )
= 1 + 122132 1 + ± 4 132
= 1 + 1441691 + | | |
By taking L.C.M on the R.H.S
We get,
cosec20 = 169+1441690 = 1691~t| 44lb9
= 313169 313169
Therefore
COSeC0=V313169 0 = y j H I
= 0 = V 3 1 3 1 3 0 =
Therefore
cosecO© = 0= V313130 = .... (2)13
Now, we know that
cosec0cosec0 = isino^p-s in fc)
sin0=iV3i3i3sin0 = -^=-13
Therefore
s in0= i3V 3 i3s in0 — .... (3)y 313
Now, we know the following trigonometric identity
cos20+sin20=1cos2 © + sin2 0 = 1
Therefore,
cos20=1— sin20cos2 0 = 1-s in2 0
Now by substituting the value of Sin0sin0 from equation (3)
We get,
2 2 COS20 = 1 —( 13V313 ) cos2 0 = 1- ^ - j = )
= 1 -1 6 9 3 1 3 1 -m
Therefore, by taking L.C.M on R.H.S
We get,
COS2© = 144313 COS2 0 = | | |
Now, by taking square root on both sides
We get,
C O S © = 12V313 COS 0 = —=V313
Therefore,
COS0= 12V313 COS 0 = - j = .... (4)
Substituting the value of sinQsin @ and COS0COS 0 from equation (3) and (4) respectively in the equation below
o ■ 1 r \ ■ 2 r \ 2 sin © C O S 02sin0 cos0 cos20 -sin 0 — — ---- 5—cos2 0-sin 0
Therefore,
2sin0COS0COS20 - S in 20 2 sin ® COS ® = 2X13V313*12V313(13V313)2-(12V313)2cos2 0-sin 0 (
2x 13 12----- , V313
13
312_ 313- 31231325313 - j j "
313
3 1 2 2 531225
Therefore
2sinOcosOcos20-sin20 2 sin ® cos ® =cos2 0-sin 0
31225 31225
19.) If COsO— 35 cos 0 = | , find the value of sinO-itano2tanOsin 0 -
ta n 0
2 tan 0
Sol.
Given: COS0=35cos0 — | .... (1)
To find the value of sin0-itano2tan0 — —2 tan ©
Now we know the following trigonometric identity
cos20 + s in 20=1cos2 0 + sin2 0 = 1
Therefore by substituting the value of COS0COS 0 from equation (1)
We get,
(35)2+sin20=1 ( | ) 2 + sin2 0 = 1
Therefore,
sin20 = 1 -(3 5 )2sin2 0 = 1 - ( | ) 2 sin20=1-(925)s in2 0 = l - ( J r ) s in20=25-925
sin2 0 = sin20 = 1625 sin2 0 = -||
Therefore by taking square root on both sides
We get,
S in0= 45s in0 = | .... (2)
Now, we know that
tan0=sin0cosOtan0 = cos ©
Therefore by substituting the value of S in 0 s in 0 and COS0cos0 from equation (2) and (1) respectively
We get,
4
tanO=4535 = 4 3 ta n 0 = -§- = | .... (4)7
Now, by substituting the value of S in0sin 0 and of ta n 0 ta n © from equation (2) and equation (4) respectively in the expression below
sin 9 - —^sin0-itane2tan0 — ----- ^ —2 tan©
We get,
1t a n 8
= 4 5 -1 4 2 X 4 3sin 0 -
sinO-itano2tan0 2 tan 0
sinO-itanG2tan0 sin 0 ~ ^ ¥ 2 tan 0 = 1620-152083
4 15 4
| CO
XCN
16 15
20 20
sin0-itane2tan0sin 0 ——^_________ tan 8
2 tan 0 = 3160 160
Therefore,
sin0-itano2tanOsin0 tanq _ . . . 3
2 tan© " J 160
cos 0 -20.) If s inO =35sin0 = | , find the value of cos0-itanG2coto —- — rp re
' 5 2 cot 0
Sol.
Given:
s in0=35s in© = | .... (1)O
cos 0 - — ^To find the value of cos0-itane2cot0 — - — —2 cot 0
Now, we know the following trigonometric identity
cos20 + s in 2O=1cos2 0 + sin2 0 = 1
Therefore by substituting the value of COS0cos 0 from equation (1)
We get,
cos20+ (35)2=1cos2 © + ( | ) 2 = 1
Therefore,
COS20 = 1 —(35)2COS2 © = 1 - ( | ) 2 COS20 = 1-925COS2 0 = 1 - ^ -
Now by taking L.C.M
We get,
COS20 = 25-925 COS2 0 = COS20 = 25-925 COS2 © = ^jjr-
Therefore, by taking square roots on both sides
COS©— 45 COS 0 — |
Therefore,
COS©= 45COS 0 = | .... (2)
Now we know that
tan© — sinOcosOtan © = sm®cos 0
Therefore by substituting the value of s inQ sin© and COS0cos© from equation (1) and (2) respectively
We get,
_3
tan©= 3545 tan. 0 = -7-_45
tan© =34tan© = | .... (3)
Also, we know that
co t0 = ItanOcot 0 = 1-tan 0
Therefore from equation (3)
We get,
COtQ= 1 34 cot 0 = y4
COt0=43COt© = | .... (4)
We get,
Now by substituting the value of COS0cos ©, tanQtan 0 and COtOcot 0 from equation (2) ,(3) and (4) respectively from the expression below
cos 0 -COS0— 1 tan© 2cotO t a n 0
2 cot 0
We get,
cos 0 -COS0— 1 tan0 2cotO t a n 8 _
-------------- 4 5 -1 3 2 X 4 3 2x!2 cot 0
COS0— 1 tan0 2C0t©C O S 0 -
t a n 0 _
2 cot 01215-2015 83
1215
2015
JS__ 15- -815 83 —
"3
C O S 0 — 7— ^ _ i
Therefore, cos0-itanG2cot0—-— = -15-?-2 cot 0 5
21.) If tan0=247tan@ = y-, find that sin0+cos©sin@ + cos0
Sol.
Given:
tan©=247tan0 = -y .... (1)
To find,
sinO+cosOsin 0 + cos 0
Now we know that tanOtan 0 is defined as follows
tan© — PerpendicularsideoppositetozOBasesideadjacenttozO
tan 0 = Perpendicular side opposite to Z© Base side adjacent to Z© (2)
Now by comparing equation (1) and (2)
We get,
Perpendicular side opposite to Z 0 Z 0 = 24
Base side adjacent to Z 0 Z 0 = 7
Therefore triangle representing Z 0 Z 0 is as shown below
Side AC is unknown and can be found by using Pythagoras theorem
Therefore,
AC2= AB2 + BC2
Now by substituting the value of unknown sides from figure
We get,
AC2= 242 +72
AC = 576 + 49
AC= 625
Now by taking square root on both sides,
We get,
AC = 25
Therefore H hy
Hypotenuse side AC = 25 .... (3)
Now we know sinOsin © is defined as follows
■ ............. . , • r\ Perpendicular side opposite to Z 0S in W - PerpendicularsideoppositetozOHypotenuseSlIl = ---------------- e?—- t-----------------------r Hypotenuse
Therefore from figure (a) and equation (3)
We get,
sin©=abac s in © =
sin©=2425sin© — .... (4)
Now we know that COS0cos 0 is defined as follows
COS0— BasesideadjacenttozOHypotenuse COS 0 Base side adjacent to Z 0 Hypotenuse
Therefore by substituting the value of sinOsin 0 and COS0COS 0 from equation (4) and (5) respectively, we get
sinO+cosOsin 0 + cos 0 = 2425+ 7 2 5 | | + ^
sinO+cosOsin 0 + cos 0 = 3125
Hence, sinO+cosOsin 0 + cos 0 = 3125
22.) If sinO=absin © = find secO+tanOsec 0 + tan © in terms of a and b.
Sol.
Given:
sin©=absin0 — | .... (1)
To find: secO+tanOsec 0 + tan 0
Now we know, SinOsin 0 is defined as follows
■ „ ................. - / " v Perpendicular side opposite to ZQS in tJ — PerpendicularsideoppositetozOHypotenuseSinfcJ = -------------- ^ — 7--------------------------r Hypotenuse
(2)
Now by comparing equation (1) and (2)
We get,
Perpendicular side opposite to Z 0 Z 0 = a
Hypotenuse = b
Therefore triangle representing Z 0 Z 0 is as shown below
Hence side BC is unknown
Now we find BC by applying Pythagoras theorem to right angled AABC A ABC
Therefore,
AC2 = AB2 +BC2
Now by substituting the value of sides AB and AC from figure (a)
We get,
b2 = a2 + BC2
Therefore,
BC2 = b2 - a2
Now by taking square root on both sides
We get,
BC= Vb2- a 2v ^ - a z’
Therefore,
Base side BC = Vb2—a2\/&2- a 2 .... (3)
Now we know COS0cos 0 is defined as follows
COS0— BasesideadjacenttozOHypotenuse COS 0 Base side adjacent to ZO Hypotenuse
Therefore from figure (a) and equation (3)
We get,
C O S © = BCAC COS 0 =
= Vb2-a2b b
C O S © = BCAC COS 0 = ^
= Vb2-a2b ^ ■■■■ (4)
Now we know, sec© = icosOsec © = —cos 0
Therefore,
sec© = bVb2-a2sec © — . ? .... (5)
Now we know, tan©=sin0cos0tan@ —C O S 0
Now by substituting the values from equation (1) and (3)
We get,
tan©=abvsq?btan© = * tan©=aVb2-a2ta n © = . f
Therefore,
tan©= aVb2-a2ta n 0 — . ° .... (6)
Now we need to find sec0+tan0sec 0 + ta n 0
Now by substituting the values of secQsec 0 and tanQtan 0 from equation (5) and (6) respectively
We get,
sec0+tan0sec 0 + tan © = bVb2-a2 + aVb2-a2 --------- h —t=—V b2 -a 2 \ /b2 -a?
sec0+tan0sec 0 + tan 0 = b+aVb2-a2 h+a .... (7)vV-a2
We get,
Sec0+tan0sec 0 + tan 0 = b+aVb+a-Vb-a /7_ T a/—VO+a-yo-a
Now by substituting the value in above expression
We get,
sec©+tan©sec 0 + tan © = Vb+aWb+aVb+a-Vb-ay o + a -y o -a
Now, Vb+a V ^ -F a present in the numerator as well as denominator of above denominator of above expression gets cancels we get,
sec0+tan0=Vb+aVb::asec0 + tan © =y / b — a
Square root is present in the numerator as well as denominator of above expression
Therefore we can place both numerator and denominator under a common square root sign
Therefore, secO+tanQsVb+aVb^asec© + tan © —y / b — a
23.) If 8tanA=158 tan >1 = 15, find sinA—cosAsin A - cos A
Sol.
Given:
8tanA=158tanA = 15
Therefore,
tanA=i58tanA = ^ .... (1)
To find:
sinA—cosAsin A - cos A
Now we know tan A is defined as follows
tanA— PerpendicularsideoppositetozABasesideadjacenttozA
, a Perpendicular side opposite to A Ata n A = — ^ -----rj— .. w ..------.... (2 )Base side adjacent to /LA v 7
Now by comparing equation (1) and (2)
We get
Perpendicular side opposite to zA /A = 15
Base side adjacent to / J K / A = 8
Therefore triangle representing angle A is as shown below
Side AC= is unknown and can be found by using Pythagoras theorem
Therefore,
AC2= AB2 + BC2
Now by substituting the value of known sides from figure (a)
We get,
AC2= 152 + 82
AC2 = 225 +64
AC = 289
Now by taking square root on both sides
We get,
AC = V289\/289
AC = 17
Therefore Hypotenuse side AC=17 .... (3)
Now we know, sin A is defined as follows
_■ * _ . a Perpendicular side opposite to / .AS in A - PerpendicularsideoppositetozAHypotenuseSin A = -------------- tj— ------------------------r Hypotenuse
Therefore from figure (a) and equation (3)
We get,
sinA=BCACsin^l —
sinA=i5i7sin>l = .... (4)
Now we know, cos A is defined as follows
COSA— BasesideadjacenttozAHypotenuse COS A Base side adjacent to Z A Hypotenuse
Therefore from figure (a) and equation (3)
We get,
cosA= a ba c c o s A =A C
COSA= 817COS A = .... (5)
Now we find the value of expression sinA— COsAsin A -co s A
Therefore by substituting the value the value of s in A s in A and COsAcos A from equation(4) and (5) respectively , we get,
sinA-cosA=i5i7-8i7sinA-cos A = j|-- jy sinA-cosA=i5-8i7
sin A-cos A = sinA—cosA=7i7sin A-cos A = ^y
Hence, sinA—COsA=7i7sin A - cos A = yy
24.) If tanO= 2021 ta n © = | y , show that 1-sinO-cosOl +sin0+cos0 =37
l-sin 0 -cos 0 _31+sin 0 + cos 0 7
Sol.Given:
tan©= 2021 t a n © -
To show that 1-sin0+cos01 +sin0+cos0 — 37 i1~sin ® +cos ® ^1+sin 0 + cos 0 7
Now we know that
tan© — PerpendicularsideoppositetozOBasesideadjacenttozO
ta n © __ Perpendicular side opposite to ZQBase side adjacent to ZQ
Therefore,
tan©= 2021 ta n 0 =
Side AC be the hypotenuse and can be found by applying Pythagoras theorem
Therefore,
AC2 = AB2 + BC2
AC2= 212 + 202
AC2 = 441 + 400
AC2 = 841
Now by taking square root on both sides
We get,
AC = V 8 4 1 \/8 4 l
AC= 29
Therefore Hypotenuse side AC= 29
Now we know, sinOsin 0 is defined as follows,
■ _ a /''k ................. . . j s-\ Perpendicular side opposite to ZQSinAfcJ- PerpendicularsideoppositetozOHypotenuseSlIl A & — -------------- 77— 7------------------------r Hypotenuse
Therefore from figure and above equation
We get,
sin© = abac sin© = 4 1 sinO= 2029 sin© =
Now we know COS0cos 0 is defined as follows
COS0— BasesideadjacenttozOHypotenuse COS 0 Base side adjacent to Z 0 Hypotenuse
Therefore from figure and above equation
We get,
COS©= ABAC COS 0 = COS©= 2129 COS 0 =A C 29
Now we need to find the value of expression l-sin0+cos0l+sin0+cos0 l-s in 0 + c o s 0 1+sin 0 + c o s 0
Therefore by substituting the value of sinOsin © and COS0cos ©from above equations, we get
1-sin0+cos01 +sin0+cos0 1-sin 0 + c o s 0 _ 1+sin 0 + c o s 0
29-20+2129 7029
29-20+21 297029
Therefore after evaluating we get,
. . _ _ _ l -s in 0 + c o s 0 _ 31-sin0+cos01+sin0+cos0 , , . ------77- 37 ■=1+sin 0 + c o s 0 7
Hence,
1-sin0+cos01 +sin0+cos0 l-s in 0 + c o s 0 1+sin 0 + c o s 0
25.) If COSecA=2cosecA = 2 , find 1tanA+sinA1+ c o s A ^ ^ - + 1 '^ £ A
Sol.
Given:
cosecA=2 cosecA — 2
To find 1 tanA + sinA1 +cosA -r~~r +tan. x\. x | cos
U ypotenuseNow cosec A = HypotenuseOppositeside —------ -— rr-Uppositesiae
Here BC is the adjacent side,
By applying Pythagoras theorem,
AC2= AB2 + BC2
4 = 1 + BC2
BC2= 3
BC = V3\/3
Now we know that
s in A = ico secA s in A = —cosecA
sinA=i2 sinA = .... (1)
tanA=ABBCtan A — 45
tanA=iV3tanA = -j= .... (2)
c o s A = b ca cc o s A = AC
cosA= V32 COS A = ^ .... (3)
Substitute all the values of sinAsin A , COsAcos A and tanAtan A from the equations(1), (2) and (3) respectively
We get.
i1tanA + sinA1+cosA — ^ - r + sin^- r = 11V3 + 12I+V32 —} -----1------
tan A 1+cos A , v3V3 1 + ~
-V3+12W3x/ 3 + ^
= 2(2+V3)2+V32(2+ V 3)
3 2+V3
= 2
Hence,
tan A 1+cos A
26.) If Z A Z A and Z B Z S are acute angles such that cos A =cos B , then show thatzA Z A = Z B LB
Sol.
Given:
Z A Z A and zB Z i? are acute angles
cos A = cos B such that Z A Z A = zB Z i?
Let us consider right angled triangle ACB
A
c B
Now since cos A = cos B
Therefore
ACAB = BCAB ^ =AC _ BCAB AB
Now observe that denominator of above equality is same that is AB
Hence ACAB = BCAB only when AC=BCAB AB J
Therefore AC=BC
We know that when two sides of triangle are equal, then opposite of the sides are also
Equal.
Therefore
We can say that
Angle opposite to side AC = angle opposite to side BC
Therefore,
z B Z 5 = z A Z A
Hence, Z A /.A = Z B /.B
27.) In a AABCAABC , right angled triangle at A, if tan C = V3-\/3 , find the value of sin B cos C + cos B sin C.
Sol.
Given:
A A B C A A B C
To find : sin B cos C + cos B sin C
The given a A A B C A A B C is as shown in figure
Side BC is unknown and can be found using Pythagoras theorem,
Therefore,
BC2= AB2 +AC2
BC2=^l32y/32 + 12 BC2 = 3+1
BC2 = 4
Now by taking square root on both sides
We get,
BC = V 4 i/4
BC= 2
Therefore Hypotenuse side BC= 2 .... (1)
Now, sin B = PerpendicularsideoppositetozBHypotenuse Perpendicular side opposite to ZB Hypotenuse
Therefore,
sinB=ACBCsin£ = ^
Now by substituting the values from equation (1) and figure
We get,
sin B = 12 \ ....(2)
Now, COS B= basesideadjacenttozBHypotenuse base side adjacent to ZB Hypotenuse
Therefore,
cos B = ABBC BC
Now substituting the value from equation
cos B= V 3 2 ^ .... (3)
Similarly
sin C = V 3 2 ^ ....(4)
Now by definition,
tanC=sinCcosCtanC =
So by evaluating
COsC=12cosC = \ .... (5)
Now, by substituting the value of sinB, cosB.sin C and cosC from equation (2) ,(3) ,(4) and(5) respectively in sinB cosC + cosB sin C
sinB cosC + cosB sin C= 12X12 + V32XV32 ^ x \ + ^ x
= M + 3 4 j + |
= 1
Hence,
sinB cosC + cosB sin C = 1
28.) State whether the following are true or false. Justify your answer.
(i) The value od tan A is always less than 1.
(ii) sec A = 125 ^ for some value of Z A Z A
(iii) cos A is the abbreviation used for the cosecant of Z A /LA.
(iv) s in 0=43s in 0 = for some angle 0 0 . Sol.
Sol.
(i) tan A < < 1
Value of tan A at 45° i.e... tan 45=1
As value os A increases to 90°
Tan A becomes infinite
So given statement is false.
(ii) sec A = 125 -y for some value of angle if
M-l
sec A = 2.4
sec A > 1
M - II
For sec A = 125 ^ we get adjacent side = 13
Subtending 9i at B.
So, given statement is true.
(iii) Cos A is the abbreviation used for cosecant of angle A.
The given statement is false.
As such cos A is the abbreviation used for cos of angle A , not as cosecant of angle A.
(iv) Cot A is the product of cot A and A
Given statement is false
cot A is a co-tangent of angle A and co-tangent of angle A = adjacentsideOpositeside
adjacent side * Sol.
Oposite side
(v) s in© =43sin© — -f for some angle 0 0 .o
Given statement is false
Since value of sinOsin 0 is less than(or) equal to one.
Here value of sinOsin 0 exceeds one,
So given statement is false.
So given statements is true.
29.) If SinO= 1213sin0 = -j| find sin20-cos202sin0cos0Xltan20 sin2 0 —cos2 0 12 sin 0 cos 0 tan2 0
Sol.
Given: S in © = l2 l3 s in 0 = -j|
To Find: sin20-cos202sin0cos0 * 1tan20sin2 0 —cos2 0 12 sin 0 cos 0 tan2 0
As shown in figure
Here BC is the adjacent side,
By applying Pythagoras theorem,
AC2=AB2+BC2
169 = 144 +BC2
BC2= 169-144
BC2= 25
BC = 5
Now we know that,
COS0— basesideadjacenttozOHypotenuse C O s Q
C O S 0 = cos0= 513 COS 0 —AC 13
base side adjacent to ZQ Hypotenuse COS0= BCAC
We also know that,
tan0=sinOcosOtan0 = :
Therefore, substituting the value of sin0sin 0 and COS0cos © from above equations
We get,
tan 0 = 1 25 tan© = ^
Now substitute all the values of sinOsin © , COsOcos 0 and tanOtan © from above
equations in sin2Q-cos2Q2sinQcosQ * 1tan2Q s*n. ®~cos ® x — \ —2 sin 0 cos 0 tan2 0
We get,
sin* 20 -cos202sin0cos0 x 1tan20 sin2 0 -c o s 2 0 2 sin 0 cos 0
Xtan2 0
2 2 2 (1213) - ( 513 ) 2 x (1213)x (5 1 3 )X 1 ( 125 )
a t r - g y :: ,KsM*) (f)!Therefore by further simplifying we get,
sin20 -cos202sin0cos0 x 1tan20 s™2. 9 cos2 ® x — = 119169 x 169120 x 251442 sin 0 cos 0 tan2 ©
119 ^ 169 25169 120 144
Therefore,
sin20 -cos202sin0cos0 x 1tan20 s*n . ® cos ®2 sin 0 cos 0X
1tan2 ©
= 5953456 5953456
Hence,
sin20-cos202sin0cos0 x 1tan20 sin2 0 —cos2 0 2 sin 0 cos 0
X 1tan2 0
= 5953456 5953456
30.) If COS©— 513 COS © = Tjj, find the value of sin20-cos202sinOcosO * 1tan20
sin2 0 —cos2 0 12 sin 0 cos 0 tan2 0
Sol.
Given: If COS0=513COS© = 7^
To find:
The value of expression sin20 -cos202sin0cos0 x ltan20 s*n. cos x2 sin 0 cos 01
tan2 0
Now we know that
COS ©COS© = basesideadjacenttoz0Hypotenusebase side adjacent to Z©
Hypotenuse (2)
Now when we compare equation (1) and (2)
We get,
Base side adjacent to Z 0 Z © = 5
Hypotenuse = 13
Therefore, Triangle representing Z0Z@ is as shown below
Perpendicular side AB is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2= AB2 + BC2
Therefore by substituting the values ogf known sides,
AB2 = 132 - 5 2
AB2= 169 -25
AB2 = 144
AB = 12 .... (3)
Now we know from figure and equation,
sin©=i2i3sin@ — .... (4)
Now we know that,
tan0=sin0cosoton© =cos 0
tan © = i25tan® — ^ .... (5)
Now w substitute all the values from equation (1), (4) and (5) in the expression below,
sin20 -cos202sin0cos0 x 1tan20 S*D. ®~cos ® x — \—2 sin 0 cos 0 tan 0
Therefore
We get,
( f y - u r :: ,
>x(s)-(i) ( f )2T h e r e fo r e b y fu r th e r s im p lify in g w e g e t,
sin20 -c o s 20 2 s in 0 c o s 0 x ita n 20 ^ 9 ~ cos2 ® x — = 1 1 9 1 6 9 X 1 6 9 1 2 0 X 2 5 1 4 42 sin 0 cos 0 tan2 ©
119 ^ 169 25169 120 144
T h e r e fo r e ,
sin20 -cos202sinOcosO x itan20 ^ 9 ~cos2 ® x — = 5953456 tS2 sin 0 cos 0 tan2 © 3456
H e n c e ,
. sin2 0 —cos2 0 w 1 _______ ___ 595sin 0-cos202sin0cos0Xitan20 - r — ----- :r— X — 5— = 59534562 sin 0 cos 0 tan2 0 3456
31.) If sec A = 178 , verify that 3-4sin2A4cos2A-3 = 3-tan2A1-3tan2A
3 -4 sin2 A __ 3 -tan2 A4 cos2 A - 3 1 -3 tan2 A
Sol.
G iv e n : s e c A = 1 7 8 ^
T o v e r ify : 3 -4 s in 2A4cos2A -3 = 3 -ta n 2A 1 -3 ta n 2A 3 4 s„in A = 3 tan ^3 4 cos2 A - 3 1 -3 tan2 A
N o w w e k n o w th a t C O S A = is e c A c o s A = —sec A
N o w , b y s u b s titu t in g th e v a lu e o f s e c A
W e g e t ,
C O S A = 817C O S A — JY
N o w w e a ls o k n o w th a t,
sin2A+cos2A= 1 sin2 A + cos2 A — 1
Therefore
• 2 2 a 1 2 2 2sin2Q-cos2Q2sinQcosQ x 1tan20 S*D. _ cos " X — = (1213) - ( 513) 2x(i2i3)x(5i3)x 1(125)2 sin 0 cos 0 tan2 0
sin2A=1-cos2Asin2 A = 1-cos2 A
= <817)2( w )2
= 225289 225289
Now by taking square root on both sides,
We get,
sinA=i5i7sin>l —
We also know tha t, tanA= sinAcosA ta n A = sin A cos A
Now by substituting the value of all the terms ,
We get,
tanA=i58tan>l = ^
Now from the expression of above equation which we want to prove:
L.H.S = 3-4sin2A4cos2A-3 4 sin \4 cos2 A - 3
Now by substituting the value of cos A ad sin A from equation (3) and (4)
We get,
3 \ 225L.H.S = 3 -42252894—64289-3 ------
4-^r-3
= 867-900256-867
= 33611 m
From expression
R.H.S = frac3-tan2A1-3tan2A/rac3-tan2 A l - 3 t a n 2 A
Now by substituting the value of tan A from above equation
We get,
R.H.S= 3 —(158) 1 -3 (1 5 8 ) ------------------2
= -3364-6116464
-61164
= 33611 m
Therefore,
We can see that,
3—4sin2A4cos2A-3 = 3-tan2A1-3tan2A 3-4 s0in2 A — 3~tan2,A4 cos2 A - 3 1 -3 tan2 A
32.) If sinO=34sin© = j , prove that Vcosec20-cot20sec2-1 = \7 3
cosec2 Q -cot2 9 -\/7sec2- l 3
Sol.
Given: s in © = 3 4 s in 0 = | .... (1)
To prove:
Vcosec2O^ot20sec=M = V/3 [ c P^c2Q-cot2 9 = ^ (2)V sec2- l 3
By definition,
sin A = PerpendicularsideoppositetozAHypotenuse Perpendicular side opposite to / .A Hypotenuse
By comparing (1) and (3)
We get,
Perpendicular side = 3 and
Hypotenuse = 4
Side BC is unknown.
So we find BC by applying Pythagoras theorem to right angled AABCAABC
Hence,
AC2 = AB2 +BC2
Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)
Therefore,
42= 32 +BC2
BC2 = 1 6 - 9
BC2= 7
BC = yfry/7
Hence, Base side BC =V7\/7 .... (3)
Now cos A = bcac4 ^
V74^ ....(4)
N o w , c o s e c A = i s i n A cosec A — - 4 - rsin A
Therefore, from fig and equation (1)
COSecA— HypotenusePerpendicular COSec.A = Hypotenuse
Perpendicular
cosecA= 43cosecA = | .... (5)
Now, similarly
secA=4\7sec A. = ..(6)
Further we also know that
COtA— cosAsinA cot A = ^ 4sm A
Therefore by substituting th values from equation (1) and (4),
We get,
COtA=V73COt A = .... (7)
Now by substituting the value of cosec A, sec A and cot A from the equations (5), (6), and(7) in the L.H.S of expression (2)
V c o s e c 2 0 ^ o t2 0 s e c ^ 1 / ^ c 20 -c o t2 0 = m n ^ lV sec2- l
n n n N (43) -(V73) (4V7) -1 ( D 2- ( ^ ) ;N B ) 2- '
= 1 6 9 -7 9 1 6 7 -1
169 9
Hence it is proved that,
Vcosec2€)-cot20sec2-1 = -.73 9 = ^V sec2- l o
33.) If SecA=178secA = ^ , verify that 3-4sin2A4cos2A-3 = 3-tan2A1-3tan2AO3 -4 sin2 A 3—tan2 A4 cos2 A - 3 1 -3 tan2 A
Sol.
Given: secA=i78secA = ^ .... (1)
To verify:
3-4sin2A4cos2A-3 = 3-tan2A1-3tan2A 3 4 sm A = 3 t a n .... (2)4 cos2 A -3 1 -3 tan2 A v ’
Now we know that sec A = icosA — —rcosA
Therefore COSA=lsecAcos A =secA
We get,
COSA=817COS A = -pf .... (3)
Similarly we can also get,
sin A= s in A = l5 l7 s inA = ■■■■(4)
An also we know that tanA= sinAcosA tan A — - ^ 4
tanA=i58tan>l = ^ .... (5)
Now from the expression of equation (2)
L.H.S: Missing close brace Missing close brace
Now by substituting the value of cos A and sin A from equation (3) and (4)
We get,
2 2 3 -4L.H.S = 3-4(1517) 4(817) -3 (# ) '
-3
- 867-900289256-867289
867-900289
256-867289
= 33611 J r . . . . (6 )611
R.H.S = 3-tan2A1-3tan2A 3—tan2 A 1-3 tan2 A
Now by substituting the value of tan A from equation (5)
We get,
R.H.S=3-(1518) 1-3(158) ,
-3364
-3364 -61164 — ^
64
= 33611 J r . . . . (7 )611
Now by comparing equation (6) and (7)
We get,
3-4sin2A4cos2A-3 = 3-tan2A1-3tan2A s„in An4 cos2 A - 3
3—tan2 A 1-3 tan2 A
sec Q cosecQ 1sec Q +cosecQ y/7
Sol.
Given: COt©=34cot@ — -74
34.) If COt0= 34cot 0 = -|, prove that sec0-cosec0sec0+cosec0 = lV7
Prove that: sec0-cosec0sec0+cosec0 — 1V7 sec ® cogec®sec 0 + c o se c 0
1_^7
Now we know that
sec0-cosec0sec0+cosec0 — sec 0 -c o se c 0 sec 0 + c o se c 0
1
Here AC is the hypotenuse and we can find that by applying Pythagoras theorem
AC2= AB2 +BC2
AC2 = 16 +9
AC2= 25
AC = 5
Similarly
sec0=ACBCsec0 — ^ sec0=53sec0 — | cosec=ACAB cosec — 4§ cosec=54AB
cosec = j
Now on substituting the values in equations we get,
sec0-cosec0sec0+cosec0 — 1V7 sec®~^osec®sec 0 + c o se c 0 y 7
Therefore,
sec0-cosec0sec0+cosec0 = 1V7 sec ® cosec®sec 0 + c o se c 0
1
Sol.
Given: 3cosO—4sin©=2cos©+sin©3 cos 0 -4 sin 0 = 2 cos 0 + sin 0
To find: tanOtan 0
We can write this as:
3cos0—4sin0=2cos0+sin03 cos 0 - 4 sin 0 = 2 cos 0 + sin 0 cos©=5sin©cos 0 = 5 sin 0
Dividing both the sides by COS0cos 0 ,
We get,
cosOcos©=5 s i n © c o s 0 = 5sin® 1=5tan©l = 5tan0 tanO=1tan© = 1cos 0 cos 0
Hence,
tanO=1tan 0 = 1
35.) If 3cosO—4sinO=2cosO+sin03 cos 0 - 4 sin 0 = 2 cos © + sin 0 .find tanOtan©
36.) If aAZA and ZP Z P are acute angles such that tan A = tan P, then show
zA = zP Z .A = Z P
Sol.
Given: A and P are acute angles tan A =tan P
Prove that: zA = zP Z A = Z P
Let us consider right angled triangle ACP
W e know t a n G —oppositesideadjacentsidetan© =
P C
tan A = pcac^77AC
tan P =a c pc^
tan A =tan P
PCAC = ACPC PC AC AC PC
PC =AC [v.-Angle opposite to equal sides are equal]
Z Z A A = Z P
Exercise 5.2: Trigonometric Ratios
Evaluate each of the following:
Q 1 . sin 45°45° sin 30°30° + cos 45°45° cos 30°30°
Solution:
Sin 45°45°sin 30°30° + cos 45°45° cos 30°30°
[1]
We know that by trigonometric ratios we have ,
sin45°=iV2sm45° =v 2
Sin30°=i2sm30°
cos45°=i\2 cos45° = A=V2 cos30°=V32cos30
Substituting the values in equation 1 , we get
W5-12+W5-V32-L . 1 + ^ . 4
Q 2 . sin 60°60° cos 30°30° + cos 60°60° sin 30°30°
Solution:
sin 60°60° cos 30°30° + cos 60°60° sin 30°30°[1]
By trigonometric ratios we have ,
sin60°=V32sm60° = ^ Sin30°=i2sm30° = \
cos30°=V32cos30° = ^ cos60°=12cos60° = \
Substituting the values in equation 1 , we get
= V32-V32 + 1 2 - 1 2 ^ ' ^ + 2 ‘ 2
= 34+ 14 4 + T = 44 4 = 14 4 4
Q 3 . cos 60°60° cos 45°45° - sin 60°60° sin 45°45°
Solution:
cos 60°60° cos 45°45° - sin 60°60° sin 45°45°[1]
We know that by trigonometric ratios we have ,
cos60°=12cos60° = \ cos45°=iV2cos45°
sin60°=V32sm60° — ^ sin45°=iV2sm45° — -±=£ \/2
Substituting the values in equation 1 , we get
12-1V2-V32-1V2-| • V3 _ J _ 2 ’
= 1-V32V2 1—\/32 /2
Q.4: sin230o+sin245o+sin260o+sin290osin230o + sm245° + sin260° + sin290c
Solution:
sin230o+sin245°+sin260o+sin290°sm230° + sin245° + sin260° + sin2 90° [1]
We know that by trigonometric ratios we have ,
Sin30°=i2sm30° = \ sin45°=iV2sm45° = V2
sin60°=V32sm60° = ^ sin90°sm90° = 1
Substituting the values in equation 1 , we get
= [12]2+ [W 2] +[V32] + 1 [ | ] 2 +n 2
L V2 J + V3 + 1
= 14 + 12 + 34 + 1 ^ + 1 + 1 + 1
= 52 —“ 2
Q 5. cos230°+cos245°+cos2600+cos2900cos2 30° + cos245° + cos2 60° + cos2 90°
Solution:
cos2300+cos2450+cos2600+cos290°cos2 30° + cos245° + cos2 60° + cos2 90° [1]
We know that by trigonometric ratios we have ,
cos30°=V32cos30° = ^ cos45°= i V2cos45° = - j -
cos60°=12cos60° = \ cos90°cos90° = 0
Substituting the values in equation 1 , we get
[V32] +[1V2] + [ 1 2 ] +0 V?2
1 2+ 1
V* + [ l ] 2 + o
34 + 12 + 14| + \ + \
321
Q 6. tan230°+tan245o+tan260°£an230o + t a n 2 45° + t a n 2 60c
Solution:
tan230o+tan245°+tan260°tan230o + t a n 2 45° + t a n 2 60°[1]
We know that by trigonometric ratios we have ,
tan30°=iV3tan30° — -7 = tan60°=V3£an60° — \/3
tan45°=1tan45° = 1
Substituting the values in equation 1 , we get
2 - 2[iV3f+[V3f+1 “ + [V 3 ]2 + 1
= i3+3+1-| + 3 + l
Q 7 .2sin230o-3cos245°+tan260o2sm230° - 3cos245° + tan260°
Solution:
2sin230°-3cos2450+tan26002sm230° - 3cos245° + tan260°
We know that by trigonometric ratios we have ,
Sin30°=i2sm30° — \ cos45°= i V2cos45° — - 7=
tan60°=V3fcm60° = \/3
Substituting the values in equation 1 , we get
= 2(i2)2-3(W i)2+(V3)22 (D 2 - 3 ( ^ ) 2 + (V3)2
= 2 ( i4 ) -3 ( i2 )+ 3 2 ( |) -3 ( i )+ 3
= 2
Q8:sin230°cos245°+4tan230°+ i2sin290°-2cos290°+ i24cos20°
sm230°cos245° + 4tan230° + |s m 290° — 2cos290° + ^cos20°
Solution:
sin230°cos245°+4tan230°+ i2sin290°-2cos290°+ i24cos20°
sm230°cos245° + 4tan230° + -|sm290o — 2cos290° + ^cos2 0°[1]
We know that by trigonometric ratios we have ,
Sin30°=i2sm30° = | cos45°=iV2cos45° = tan30°=iV3tan30°
sin90°sm90° = 1
cos90°cos90° = 0
cos0°cos0° = 1
Substituting the values in equation 1 , we get
2[0]2 + £ [ l ]2
= 18 + 43 + 1 2 + 1 2 4 i + | + i + ±
= 4824 | | = 2
Q 9 .4(sin460°+cos4300)-3(tan2600-tan2450)+5cos245°4 (sm 460° + cos430°) — 3 ( t a n 2 6 0° — t a n 2 45°) + 5cos245°
Solution:
4(sin460°+cos4300)-3(tan2600-tan2450)+5cos245°4 (sin460° + cos430°) — 3 { t a n 2 60° — t a n 2 45°) + 5cos245°[1]
We know that by trigonometric ratios we have ,
sin60°=V32sm60° = ^ cos45°=iV2cos45° = -4-2 ^2
tan60°=V3fan60° = a/3 cos30°=V32cos30° = ^
Substituting the values in equation 1 , we get
4([V32] +[V32] )-3 (3 )-1 2+5[iV2] 4 r + 5 i_V2
= 4-1816-6+524 • -ff — 6 + I
= 14-6+52 | - 6 + |
= 1 4 2 -6 ^ - 6 = 7 - 6 = 1
Q 10. (cosec245°sec2300)(sin2300+4cot2450-sec260°)(cosec2 45° sec2 30°) (s in2 30° + 4cot245° — sec2 60°)
Solution:
(cosec245°sec2300)(sin2300+4cot2450-se c260°)(cosec2 45° sec2 30°) (s in2 30° + 4cot245° — sec2 60°)
[1]
We know that by trigonometric ratios we have ,
cosec45°=V2cosec45° = y / 2 sec30°=2V3sec30° =
Sin30°=i2sin30° = \ cot45°cot45° = 1
sec60°sec60° = 2
Substituting the values in equation 1 , we get
- 2([V2] .[*& ] ) ( [ i2f +4(1)(2)2) [^2] 2
v^J+ 4(1) (2)2)
= 3-43-143 • | • 14
Q11. cosec3300cos60°tan3450sin2900sec2450cot300cosec3 30° cos60° tan3 45° sin2 90° sec2 45° cot30°
Solution:
Given,
= cosec330°cos600tan3450sin2900sec2450cot30° cosec3 30° cos60° tan3 45° sin2 90° sec2 45° cot30°
= 23(i2 )(13)(12)(V22)(V3)23( i ) ( l 3) ( l 2) ( ^ 2)(V 3 )
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(cos0°+sin450+sin300)(sin900-cos450+cos60°)(1 + 1V2 + 1V2 )(1—1V2 + 1V2 )
( 32 + 1V2 )( 32 — 1V2 ) (( 32 )2—(1 V2 )2) 94 — 12 74
(cosO° + sm45° + sm30°)(sm90° — cos45° + cos60°)
f 1 + 75 + + 75^
( ( 3\2 _ Z J_^2\ 9 _ 1 7 U 2 ^ ' ^/2 ' ' 4 2 4
Q14. sin30°-sin90°+2cos0°tan30°tan60° sin30°—sin90°+2cos0°tan30°tan60°
Solution:
Given,
sin30o—sin90°+2cos0°tan30°tan60°
_ Osin30°-sin900+2cos00tan300tan60° 12-I+ 21V3W3 32 ^
Q15. 4cot230° + 1sin260°—COS245°— } ------h ~ r 4 ----------COS245°cot2 30° sin2 60°
Solution:
Given,
4cot230° + 1 sin260° -C O S 24 5 ° = 4(V3)2 + 1 (V32)2—( 1<2 ) 2= 43 + 43 - 12 = 16-36 = 136
cot2 30c +
(V3)2+
sin2 60° 1
( f ) !4 1 4 _ 1 3 ' 3 216-3
- C O S 2 45°
136
Q16.4(sin4300+cos260°)-3(cos2450-sin2900)-sin26004(sm430° + cos26 0 ° ) — 3(cos245° — sin2 90°) — sin2 60°
Solution:
Given,
4(sin430°+cos2600)-3(cos2450-sin2900)-sin260°=4(( 12 )4+( 12 )2)-3(( 1V2 )2-1 )-
( V32 )2=4( 116 + 1 4 )+ 32 — 34 = 84 =2
4(sm430° + cos26 0 ° ) — 3(cos245° — sin2 90°) — sin2Q0° = 4 ( ( i)4 + (D 2) - 3 ( ( ^ ) 2 - l ) - ( # ) 2
= 4 ( ^ + 1) + ! - !_ 8 _ o
Q17. tan260°+4cos245°+3sec230°+5cos290°cosec30°+sec60°-cot230'
tan2 60°+4cos245°+3sec2 30°+5 cos2 90° cosec30°+sec60°—cot2 30°
Solution:
Given,
tan260°+4cos2450+3sec2300+5cos2900cosec300+sec60°-cot2300 = (V3)2+4(iV2)2+3(2V3)2+5(0)2+2-
tan260°+4cos245°+3sec230°+5cos290° cosec30°+sec60°—coi2 30°
_ (a )2+ 4 (^ )2+ 3 (^ )2+5(0)
2 + 2 -(V 3 )2
= 3 + 2 + 4 (V3)2=3+2+4=9= 9
Q18. sin30°sin45° + tan45°sec60°—sin60°cot45°—cos30°sin90°smA5
I tan45° sin60°”l~ sec60° coM5°
Solution:
cos30°sm90°
Given,
sin30°sin45o + tan45osec60°—sin60°cot45°—cos30°sin90° — 121V2 + 1 2 —V32I — V32I — V22 + 12—V32
sin30° 1 sinAh0
tan45° sm 60° cos30'sec60° cot 45° sin90
y3
J _ 1 2 1 1V2
V32 = V2+1-2V32 2
, 1_ ^3_ v f2 ' 2 2 2
^+1-2^3
1 i2 ' 2
Q 19. tan45°cosec30° + sec60°cot45° + ssin90°2cos0 tan45° + sec60'cot45'
+ ssin90'2 cos0 °cosec30
Solution:
Given,
tan45°cosec30° + sec60°cot45° + ssin90°2cos0° = 12 + 21-5(1 )2(1) = 5 2 -5 2 =0tan45° , sec60° , ssin90°
cosec30° cot45° 2cos0°
_ 1 , 2 5(1)2 1 2(1)5 _ 5 2 2
= 0
Q20.2sin3x=V32sm3cc = y/%
Solution:
Given,
2sin3x — y/3
= > sin3x — y
— > sin3x = sin60° = > 3 x = 60°
2sin3x=V3=>sin3x=V32=>sin3x=sin60°=>3x=60°=>x=20o=> x = 20°
Q 21)2s in x2= 1 ,x= ? 2sm | = 1, x =?
Solution:
sinx2 = i2 sm | = \ sinx2=sin30°sm§ = sin30° x2=30°f = 30°
x = 60°
Q22) V3sinx=cosxv^3sma: = cosx
Solution:
V3tanx=1\/3 tanx = 1 tanx=iV3tanx = .’.tanx=tan45°.\ tanx = £an45°
x = 45°
Q23) Tan x = sin 45° cos 45° + sin 30°
Solution:
T anx= 1V2.1V2+i2[vsin45°= iV2COs45°= iV2sin30°= 12]
Tanx — [y sm45° — cos45° — sm30° = -|] T a n x = i2 + i 2
Tanx = ^
Tan x = 1
Tan x = 45°
x = 45°
Q24) V3Tan2x=cos60°+sin450cos45°\/3 Tan2x = cos60° + sm45°cos45°
Solution:
V3T an2x= 1 2 + 1V2 . 1V2 [vcos60°= i2sin45°=cos45°= 1 V2 ]
y/3 Tan2x — I + [v cos60° — sin45° — cos45° — ]
V3T an2x=iV3=>tan2x=tan30°v/3 Tan2x = -j= ^ tan2x = tan30°
2x = 30°
x = 15°
Q25) cos2x=cos600cos300+sin60°sin300cos2x — cos60° cos30° + sm60°sm30°
Solution:
cos2x= 1 2 . V3 2 +V3 2 . 12 [vcos60°=sin30°= I2sin60°=cos30°= V3 2 ]
cos2x — I . ^ + "T"- \ [’•’ cos60° — sm30° — -|sm60° — cos30° —cos2x=2.V34cos2a: — 2 .^ cos2x=l32cos2x — ^ cos2x=cos30°cos2a; — cos30°
2x=30°2a: = 30° x=15°:c = 15°
Q26) I f 6 = 30°, verifylf0=3O°,verify(i)T an20=2Tanei-tan2e(i)Tan20 = -
Solution:
T an20=2Tanei-tan2e ..... (\)Tan20 = .......(i)
Substitute 0=30°0 — 30° in equation (i)
LHS = Tan 60° = V3V3
_ 2__LR H S = 2T an30°1 + (T an30°)2 = 2 -1 Vi 1 - ( 1V2)2 = V 3 2Tan30° = -------- = ^ / 3
l+ (T a n 3 0°)2 l - ( - L ) 2V 2
Therefore, LHS = RHS
(ii) sin0=2tanei-tan 2%sin0 = 2tane0n1 —tan*0
Substitute 0=3O°0 = 30°
S in 6 0 0=2tan30°(1-tan30°)2s in 6 0 0 = 2ton30°n .(1—tan 30°)2
— 2 —1-=>V32 = 2.iVi1+(iVi)2 ^ — ------
2 !+ ( ^ ) 2
= > V 3 _ 2 32 y/3 m 4
— .V32 = 2V3.34 => V32 = V32 ^ ~2~ — ~2~
Therefore, LHS = RHS.
(iii) COS20=1-tan2ei+tan2ecos20 1— tan2 6
1 +tan 2 0
Substitute 0=3O°0 = 30°
LHS = cosec0cosec0 RHS = 1-tan201+tan20 1—tan2 6
1+tan 2 6
= cos 2(30°) = 1 -ta n 230°1 +tan230° =
Cos 60°= 12 — 1 —(1V2)21 +(1V2)2 — 22 12 — 12 =1+( ^ ) 2
1—ton230°
l + i a n 230°
22_ _ X1 — 22
Therefore, LHS = RHS
(iv) c o s 36=4c o s36 -3 c o s6cos30 = 4 c o s 3 6 — 3 c o s 0
Solution:
LHS = cos30cos30
Substitute 0=3O°0 = 30°
= cos 3 (30°) = cos 90°
= 0
RHS = 4cos30-3cos04cos30 - 3cos6
= 4cos330°-3cos30°4cos330° - 3cos30°
= 4 (V 32 )3- 3 . V 3 2 4 ( ^ ) 3 - 3 . ^
= 3 -V 3 2 -3 .V 3 2 3 .4 -3 .4
= 0
Therefore, LHS = RHS.
Q27) If A = B = 60°. Verify (i) Cos (A - B) = Cos A Cos B + Sin A Sin B
Solution:
Cos (A - B) = Cos A Cos B + Sin A Sin B....... (i)
Substitute A and B in (i)
=>cos (60° - 60°) = cos 60° cos 60° + sin 60° sin 60°
=>COS 0° = (12)2+(V32)2( i ) 2 + ( ^ ) 2
=>1= 14 + 3 4 | + |
= >1 = 1Therefore, LHS = RHS
(ii) Substitute A and B in (i)
=>sin (60° - 60°) = sin 60° cos 60° - cos 60° sin 60°
=> sin 0° = 0
=>0 = 0
Therefore, LHS = RHS
(iii) T a n (A -B )= T an A -T an B 1 + T an A T an B T an (4 - B) = anATanB
A = 60°, B = 60° we get,
T a n ( 6 0 0 —6 0 ° ) = T a n 6 0 °-T a n 6 0 °1 + T a n 6 0 °T a n 6 0 °T £ m ( 6 0 ° - 6 0 ° ) = rffln6°°
Tan 0° = 0
0 = 0
Therefore, LHS = RHS
Q28 ) If A = 30°, B = 60° verify:
(i) Sin (A + B) = Sin A Cos B + Cos A Sin B
Solution:
A = 30°, B = 60° we get
Sin (30° + 60°) = Sin 30° Cos 60° + Cos 30° Sin 60°
Sin (90°)= 12.12 + V32.V32-|.-| +
Sin (90°) = 1 => 1 = 1
Therefore, LHS = RHS
Tan60°) Tan60°
(ii) Cos (A + B) = Cos A Cos B - Sin A Sin B
A = 30°, B = 60° we get
Cos (30° + 60°) = Cos 30° Cos 60° - Sin 30° Sin 60°
Cos (90°)= 12 .V 32 -V 32 .12^ \
0 = 0
Therefore, LHS = RHS
Q29. If sin(A+B) = 1 and cos(A-B) = 1, 0°<A+B<90°0° < A + B < 90°, A>B find A and B.
Sol:
Given,
sin(A+B) = 1 this can be written as sin(A+B) = sin(90°)sm(90°)
cos(A-B) = 1 this can be written as cos(A-B) = COS(0°)cos(0°)
=> A + B = 90°90°
A - B = 0°0°
2A = 90°90°
. Qf1°A = 90°2^-
A = 45°45°
Substitute A value in A - B = 0°0°
45°45° - B = 0°0°
B =45°45°
Hence, the value of A = 45°45° and B =45°45°
Q30. If tan(A-B) = 1V3 4 * and tan(A+B) = V3a/ 3, 0°<A+B<90°0° < A + B < 90°,
A>B find A and B
Solution:
Given,
tan(A-B) = 1V3-^
A - B =tan 1 ( 1V3 )tan~1(-^)
A - B = 30°30° ------1
tan(A+B) =
A + B = tan_1V3tan_l\/3
A + B = 60°60° -------2
Solve equations 1 and 2
A + B = 30°30°
A - B = 60°60°
2A = 90°90°
A = 90°2 ^
A = 45°45°
Substitute the value of A in equation 1
45°45° + B = 30°30°
B = 30°30° - 45°45°
B = 15°15°
The value of A = 45°45° and B = 15°15°
Q31. If sin(A-B) = 12 \ and cos(A+B) = 12-|, 0°<A+B<90°0° < A + B < 90°, A<B
find A and B.
Solution:
Given,
sin(A-B) = 12-|
A - B = sin_1( i2)sm_ l(-|)
A - B = 30°30° -------1
cos(A+B) = 12-|
A + B = COS-1 (12 )cos_1( | )
A + B = 60°60° -------2
Solve equations 1 and 2
A + B = 60°60°
A - B = 30°30°
2A = 90°90°
A = 90°2 ^
A = 45°45°
Substitute the value of A in equation 2
45°45° + B = 60°60°
B = 60°60° - 45°45°
B = 15°15°
The value of A = 45°45° and B = 15° 15°
Q32. In a A A ABC right angled triangle at B, ZA=Z.CZA ZC.Find the values of:
1. sinAcosC + cosAsinC
Solution:
since, it is given as zA=zC/.A — Z.C
the value of A and C is 45°45°, the value of angle B is 90°90°
because the sum of angles of triangle is 180°180°
=> sin(45045°)cos(45°450) + cos(45045°)sin(45045°)
=> x - i ) + ( i ^ x ^ X - i x -3 j)
=> 12-| + 12 i
= > 1
The value of sinAcosC + cosAsinC is 1
2. sinAsinB + cosAcosB
Solution:
since, it is given as zA=zCZ_A = Z.C
the value of A and C is 45°45°, the value of angle B is 90°90°
because the sum of angles of triangle is 180°180°
=> sin(45°45o)sin(90°90o) + cos(45°45o)sin(90°90o)
=> WS-^(O)
=> 1 ^ ^ + o
= > 'f2T2
The value of sinAsinB + cosAcosB is 1V2 -4=y/2
- /3Q33. Find the acute angle A and B, if sin(A+2B) = V32 and cos(A+4B) = 0, A>B.
Solution:
Given,
sin(A+2B) = V 3 2 ^
A + 2B = Sin-1 V32 sin~1 -
A + 2B = 60°60° ------- 1
Cos(A+4B) = 0
A + 4B = s i r r 1(90 )sm _ l (90)
A + 4B = 90°90° ------- 2
Solve equations 1 and 2
A + 2B = 60°60°
A + 4B = 90°90°
(-) (-) (-)
-2B = -30°30°
2B = 30°30°
B = 30°2
B = 15°15°
Substitute B value in eq 2
A + 4B = 90°90°
A + 4(15°15°) = 90°90°
A + 60°60° = 90°90°
A = 90°90° - 60°60°
A = 30°30°
The value of A = 30°30° and B = 15°15°
Q 34. In AP QRAPQR, right angled at Q, PQ = 3 cm and PR = 6 cm. Determine Z Z P
and Z Z R.
Solution:
Given,
In APQRAPQR, right angled at Q, PQ = 3 cm and PR = 6 cm
By Pythagoras theorem,
PR2=PQ2+QR2=>62=32+QR2=>QR2=36-9=>QR=V27=>QR=3V3PR2 = PQ2 + QR2 = > 62 = 32 + QR2 = > QR2 = 36 -9
= > QR = y/27 —> QR — 3 /3
sin R = 36 = I2=sin30°| — — sm30°
zR=30°Zi? = 30°
As we know, Sum of angles in a triangle = 180
zP +zQ+zR= 180°=>zP +90°+30°=180°=>zP=180°-120°=>zP =60°Z P + Z Q + Z i* = 180°= > Z P + 9 0 ° + 30° = 180°= > Z P = 180°-120°= > Z P = 60°
Therefore, zR=30°ZP = 30°
And, z P = 6 0 °Z P = 60°
Q35. If sin(A - B) = sin A cos B - cos A sin B and cos (A - B) = cos A cos B + sin A sin B, find the values of sin 15 and cos 15.
Solution:
Given,
sin(A - B) = sin A cos B - cos A sin B
And, cos (A - B) = cos A cos B + sin A sin B
We need to find, sin 15 and cos 15.
Let A = 45 and B = 30
sin 15 = sin (45- 30) = sin 45 cos 30 - cos 45 sin 30
= = ( 7 2 X X_ _ _ _ _ _ V3-1=(1V2XV32)-(1V2X12)=V3-12V2—
cos 15 = cos (45- 30) = cos 45 cos 30 - sin 45 sin 30
= = ( ^ 2 X ^ + X
=(1V2XV32)+(1V2X12)=V3+12V2=
Q36. In a right triangle ABC, right angled at C, if ZB=60°ZP = 60° and AB=15 units. Find the remaining angles and sides.
Solution:
S in60°=x15 V32 = x15X=15V32UnitSCOS60°=x15 12 = x15X =152X =7.5un itS
sm60° = f5V3 _ x2 ~ 152 15
X
cosQO0 = §___ X
2 15
x = 7.5units
Q37. In AAABC is a right triangle such that Z C = 9 0 °ZC = 90°, Z A = 4 5 °ZA = 45° and BC = 7 units. Find the remaining angles and sides.
Solution:
C
B
7
A
X
Here, zC = 9 0°Z C = 90° and zA =45 °Z A = 45°
We know that,
z A + z B + z C Z A + Z B + Z C = 180°180°
=> 45°45° + 90°90° + z C Z C = 180°180°
=> 135°135° + z C Z C = 180°180°
=> z C Z C = 180°180° - 135°135°
=> Z.CZC = 45°45°
The value of the remaining angle C is 45°45°
Now, we need to find the sides x and y
here,
cos(45) = BCAB
1V2 -h= = 7 y -y/2 y y
y = 7V27\/2 units
sin (45) = ACAB
lV2 -i= = xy —2/
1V2 —7= = x7V2V2
x7 2
x = 7’& 5 ^
x = 7 units
the value of x = 7 units and y = 7V27-\/2 units
Q 38 . In a rectangle ABCD , AB = 20 cm , ZZBAC = 60°60° , calculate side BC and diagonals AC and BD .
Solution:
Let AC = x cm and CB = y cm
Since , COSQcosO = basehypotenuse t— —hypotenuse
Therefore , COs60°=20xcos60° = —
=> 12 —20x=> ^ [since,COs60°=12cos60° = -|]
=>=> x= 40 cm = AC
Similarly BD = 40 cm
Now,
Since , SinGsmf? = perpendicularhypotenuse PeJ'Pen icularhypotenuse
Therefore , sin60°=BCACsm60° = A C
=>V32 = y40=^ ^ ^ =>y=40V32^> y = ^
=>y=20V3^> y = 20a/3 cm .
Q39:lf A & B are acute angles such that tanA=1/2 tanB=1/3 and tan(A+B)=tanA+tanB1-tanAtanB tanA+tanB
1—tanA tanB, find A+B.
Solution:
1_|_I i± l 1T an(A+B)= 12 + 131—12.13 Tan(A + B ) — 2 1 31 3+2656 = —|---- 5656= -jr-
2' 3 6 6Tan(A+B)=56><65Tan{A + B) = § x § (A+B)=TarT1(1)(A + B) = T a n ^ i l )
(A+B) = 45°
Q40: Prove that: (V3-1)(3-C O t30o)=tan360-2s in60°( V 3 — 1)(3 — cot30°) = tan3 60 — 2sm60°
oOCO-wou
(0c<
ooCO"o
AIICOI
+ICO
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oCOo0
1CO
Jco
rH+
ICO
I co
+ +\co \co
<M
II
I
= 2V
32\/3
R.H
.S =
> ta
n360
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n60°
£an3
60 —
2sm
60'o
XCM
'9xCMIco
I
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3-V
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Exercise 5.3: Trigonometric Ratios
Q.1) Evaluate the following : (i) sin20cos70
Sol ( i) : Given that, sin20cos70' ’ ’ cos 70
Since sin (90 - 0 0 ) = cos0@
sir, on sini90—70)=>=> sin20cos70-^|^= sin(90-70)cos70— 7n
=>=>- sin20cos70 sm 20 _cosjq ~ cos70cos70 cos70
cos70
=>=>- sin20cos70 sin2 0cos70 =1
Therefore sin20cos70 stnl® = 1cos 70
(ii) cos19sin71 cos!9sin71
Soln.(ii): Given that, cosl9sin7l cos!9sin71
=>=>- cos19sin71 cos\9 _ sin71
cos( 90—71)cos(90-71)sin71 sin71
=>=£>cos19sin71 cos!9sin71
= sin71sin71 sm71sin71
cos19sin71 cos!9sin71 = 1
Since cos(9O-0@)= sin 0 0
Therefore cosl9sin7l cos^ = 1
(Mi) sin21cos69 sin21cos69
Soln.(iii): Given that, sin2lcos69 sin21co«69
Since (90-00) = cos00
=>=y>sin21cos69 ^ gQ = sin(90-69)cos69
sin21cos69 sin 21 cos 69
= cos69cos69 cos 69 cos 69
sin21cos69 sin21cos69 =1
(iv) tan10cot80 tanlOcot80
Soln.(iv): We are given that, tanl0cot80 tanlOcot80
Since tan(9O-00) = cot00
-tan10cot80-^^ =tan(90-80)cot80 ian(90—80) cot80
tan10cot80
cot80cot80 cot80cot80
tanlO _ cot80
=>=>- tan10cot80 .on = 1cotm
T h e r e fo r e tan l0co t80tanlOcotSO = 1
(v) sec11cosec79 sec' ' cosecTd
Soln.(v):
Given that, sec11cosec79
Since sec(9O-O0)=cosec©0
=>=£> sed 1 cosec79 secllcosec79
sec(90-79)cosec79 sec(90—79) cosec79
=>=>- sed 1 cosec79 secllcosec79 = cosec79cosec79 cosec79
cosec79
=>=>- sed 1 cosec79 secllcosec79 = 1
Therefore sec11cosec79 aeclL = 1cosec79
Q.2: EVALUATE THE FOLLOWING :
(i) (sin49°cos41° \ / sin49° \ • V c o s4 1 °y
2+ (cos41°sin49° \ ( cos41° \
• \ sm 49° )
2
Soln.(i):
2 / . Q \ 2 2 / o \ 2We have to find: (sin49°cos4l°) ( ) + (cos4l°sin49°) ( cost\l )
v ' y cos41 ) v ' \ sin 4 9 ° J
Since sec70°cosec20° sec7®° + sin59°cos31° sin^ ° sin(90°90° - 0 0 ) = cos©@ and cos(90°COo g C ZU COStj J.
90° - 0 0 ) = sin0©
So
( sin(90°-41 °)cos41 ° ) ( ) + (cos(90°-49°)sin49°) ( )
( cos41°cos41°) + ( Sin49°Sin49°) ( g ^ )
= 1+ 1 = 2
So value of (sin49°cos41°) + (cos41°sin49°) is 2
(ii) cos48°cos48° - sin42°sm42°
Soln.(ii)
We have to find: COs48°cos48° - Sin42°sm42°
Since cos(9O°-09O° — 0 ) = sin00.So
cos48°cos48° -sin42°sm42° = cos(90°-420)-sin42°cos(900 - 42°) - sm42°
=sin42°sm42° - sin42°sin42° =0
So value of cos48°cos48° - Sin42°sm42° is 0
(iii) c o l4 0 - te n 5 0 -1 2 (c o ,3 5 -s in 5 5 - )^ “ l ( ^ § )
Soln.(iii)
We have to find:
cot40°tan50°— 12 ( cos35°sin55°) \ ( CQS )' ' tanoO i \ stnoo )
Since cot(9O°-09O° — 0 ) = tan0@ and cos(9O°-09O° — @)=sin©@
cot40°tan50°— 12 ( cos35°sin55°) \ ( ^*55° ) = cot(90°-50°)tan50° - 12(cos(90°-55°)sin55°)
coi(90°—50°)t c m 5 0 °
1 / cos(90°—55°) \2 \ sm 55° )
= tan50°tan50° -1 2 ( sin55°sin55°) 1. / sm 55° \ 2 ^ sm 55° y
= 1-121 - \ = 12±
So value Of C o t4 0 -ta n 5 0 -1 2 (co s 3 5 -s in 5 5 -)^ - l ( ^ g p ) is 12±
(iv)(sin27-cos63')2 ( ^ P ) 2 - (cos63-sln2r)2 ( t § - )
Soln(iv)
We have to find: (sin2Tcos63-)2( ^ ) 2 - (c o s e a w ^ ^ ) 2
Since sin (9O°-09O° — 0)= cos00 and cos (90°-©90° — 0)=sinO0
2 2 2 2 2 (sin27“cos63°) - (cos63'sin2r) = ( sin(90'-63’ )cos63-)
- ( ooS(90.-27> „ 2^ 2( ^ M ) 2
= ( cos63-oos63-)2 ( ^ | ; ) 2 _ ( sln2rsln2r )2 ( ^ ) 2 = 1-1 =0
Sovalueof(sin27-coS63-)2 ( ^ ) 2 -(cos63-sin2r ) 2 ( ^ | ; ) 2 is 0
(v)tan35°cot550 + co t7 8 °ta n 1 2 °-1 ^ !^ - +' ' cot55 tanl2 — 1
Soln.(v)
We have to find:
tan35°cot55° + cot78°tan12°— 1 ^ 77^ + f°*78° - 1cot55 tanl2°
/ sin(90°—63°) Y cos63°
Since tan (9O°-09O° — ©)= cot0© and cot (90°-©90° — 0)=tanO0 =1
So value of tan35oC0t55° + C0t78otan12°is1 + f°*7f 0 251catbo tanl2
(vi) sec70°cosec20° + sin59°cos31° sec no ' 7 cosec20 + sin59°
cos31°
Soln.(vi)
We have to find: sec70°cosec200 + sin590cos31° + sm 59°cos31°
Since sec70°cosec20° + sin59°cos310 sec7 0 + and sec(90°—090° — @)=cosec©@cosec20 cosZl ' '
So
2 2sec70°cosec20° + sin59°cos31 ° aec7°° + =( sec(90°-20°)cosec20°) f SeC ° ~^0° ) -
cosec20 cos31° ' 7 \ cosec20° /
( * „ ( 9 0 - - ~ 0 2( ^ ^ ) 2
= cosec200cosec200 + cos310cos31°-£2^ ^ + CQS ° =1+1 =2cosec20 ' cos31
So value of sec700cosec20° + sin590cos31° sec7°° + sm ° is 2COocCZU COdOl
(vii) cosec310-sec590cosec3l0 — sec59°.
Soln(vii)
We have to find: COSec31°-sec59°cosec3l° — sec59°
Since COSec(9O°-0)cosec(9O° — 0)=sec00.So
= cosec31°-sec590cosec3l° - sec59°
= cosec(90°-59°)-sec59cosec (90° - 59°) - sec59 =sec59°-sec59° sec59° - sec59° =0
So value of cosec31°-sec590cosec3l° — sec59° is 0
(viii)(sin72°+cos18°)(sm72° + cosl8°) (sin72°-cos18°)(sm72° — cosl8°)
Soln.(viii)
We have to find: (sin72°+cos18°)(sm72° + cosl8°) (sin72°-cos18°)(sm72° — cosl8°)
Since sin(9O°-0)sm(9O° — 0)=cos0©,So
(sin72°+cos18°)(sm72° + cosl8°) (sin72°-cos18°)(sm72° - cosl8°)= (sin72°)2 (sin72°) 2-(cos 18°)2 (cos 18°)2
=[sin(90°-1 8°)]2- ( cos1 8°)2[sm (90° - 18°)]2 - (cosl8°)2
=(cos1 8°)2(cos18°)2- (cos1 8°)2(cos18°)2
=cos218°-cos218°cos218° - cos218° =0
So value of (sin72°+cos18°)(sm72° + cosl8°) (sin72°-cos18°)(sm72° — cosl8°) is 0.
(ix)sin35°sin550sm350sm550-cos350cos550cos350cos55°
Soln(ix)
We find :
Sin350sin550sm350sm550-cos350cos55°cos350cos550
Since sin(9O°-0)sm(9O° — 0)=cos00 and COS(9O°-0)cos(9O° — 0)=sinO0
Sin350sin550sm350sm550-cos350cos550cos350cos55°= sin(90°-55°)sin55° sm(90° — 55°)sm550-cos(900-550)cos550cos(90° — 55°)cos55° =1-1 =0
So value of sin35°sin550sm350sm550-cos350cos550cos350cos55° is 0
(x) tan480tan230tan420tan670£an48°tcm230£an420£an670
Soln.(x)
We have to find tan48°tan230tan420tan670tan480£an230£an420£an67°
Since tan(90°-©)£an(90° — @)=cot©@. So
tan480tan230tan420tan670£an480£an230tara42°£cm670 = tan(90°-42o)tan(90o-67o)tan42otan67°tan (90° - 42°) tan (90° - 67°) tan42°tan67°
=C0t42°C0t67otan42otan67°coM2occtf67o£an42o££m67o
=(tan670cot670)(tan420cot420)(£an67°co£670) (£an42°cot42°) =1*1 =1
So value of tan480tan230tan420tan670£an480£<m23°£cm420tan670 is 1
(xi)sec50osin40o+cos40ocosec50osec50osm40° + cos40°cosec50°
Soln.(xi)
We have to find sec500sin400+cos400cosec50°sec500sm400 + cos40°cosec50°
Since COS(9O°-0)cos(9O° — @)=sin0@,sec(9O°-0)sec(9O0 — 0)=cosec0© and sin00.cosec©0=1. So
sec50°sin400+cos400cosec500sec50°sm400 4- cos40°cosec50° = sec(90°-40o)sin40°sec(90o - 40°)sm40° +cos(90°-50°)cosec50° cos (90° — 50°)cosec50° =1+1 =2
So value of sec50osin40o+cos40ocosec50osec50°sm400 + cos40°cosec50° is 2.
Q.3) Express cos75°750+cot75°750 in terms of angle between 0° and 30°.
Soln. 3 :
Given that: cos75°750+cot75°750
=cos75075°+cot75°750
= cos(90°-15°)+cot(90°-15°)cos (90° - 15°) + cot (90° - 15°)
=sin 15°l5°+tan 15°15°
Hence the correct answer is sin 15°15°+tan 15°15°
Q.4) If sin3A = cos(A - 26°), where 3A is an acute angle, find the value of A.
Soln.4:
We are given 3A is an acute angle
We have: sin3A=cos(A-26°26°)
=>=^sin3A=sin(90°90o-(A-26o26°))
=>=>- sin3A=sin(116°116°-A)
=>= 3A=116°116°-A
=>=^4A=116°1160
=>=^A=29°29°
Hence thecorrect answer is 29°29°
Q.5)lf A, B, C are the interior angles of a triangle ABC, prove that,
(i) tan(c+A2)=cotB2tan = cot^
(ii) sin(B+c2)=cosA2sin = cos^
Soln.5:
(i)We have to prove: tan(c+A2)=COtB2tcm ( ^ y ^ ) = cot-f
Since we know that in triangle ABC
A+B+C=180
=>=>C+A=180°180°-B
=> > C+A2 —90°— B2 ~ = 90° - f
=>^tanc+A2=tan(90°B2)£an^^ = tan (90°-j)
=>=^tan(c+A2)=cotB2*an = cotf
Hence proved
(ii)We have to prove : sin(B+C2)=COSA2sin —
Since we know that in triangle ABC
A+B+C=180
=>=>B+C=180°180°-A
B+C2 —90°—A2 — = 90° - 4=>^sinB+C2=sin(90oA 2 )s m ^ 1 = sin (90°
sin(B+c2)=cosA2sm = cos4
Hence proved
Q.6)Prove that:
(i)tan20o20otan35o35otan45o45otan55o55otan70°70'
(ii) sin48°48° .sec48°48° +cos48°48° .cosec42°42° =
(iii) sin70°cos20° + cosec20°sec70°~2COS70°.COSGC20°:
+ coseg01 _ 2 co« 7 0 ° . cosec 20° = 0cos20 sec70
C O S
= 1
0
Soln.6:
(i) Therefore
tan20o20otan35o35otan45o45otan55o55otan70°70o
=tan(90°-70°)tan (90°-70°) tan(90°-55°)tan (90°-55°) tan45°45°tan55° 55°tan70°70°
=cot700700cot550550tan45°450tan550550tan700700
= (tan700cot700)(tan550cot550)tan450(£an700co£700) (tan&b0cotbb0) tan45° =1x1x1 =1
Hence proved
(ii) We will simplify the left hand side
sin48°48° .sec48°48° +cos48°48° .cosec42°42° =sin48°48°. sec(90°-48°) sec (90° - 48°)cos48°480 .cosec(90°-48°)cosec (90° - 48°)
=sin480480.cos480480+cos480480.sin48048° =1+1 =2
Hence proved
(iii) We have,sin70°cos20° + cosec20°sec70°—2cos70°.COSec20°=0
singl + _ 2cos70°. cosec20° = 0cos20 sec70
So we will calculate left hand side
sin70°cos20° + cosec20°sec70° -2cOS70°.COSec20°=0
(iv) c o s 8 0°s in io °+ c o s 5 9 °.c o s e c 3 1 °= 2 ^ g ^ + cos59°.cosec31° = 2
svn^ + cose^_ _ 2cos70°. cosec20° = 0=cos20 sec70
sin70°cos20° + cos70°sin20° -2COS7 0°. COSec(90°-70°)
sin70° cosec20°sec70°
- - 2cos70°.cosec(90° - 70°)cos 20' sin20
= sin(90°-20°)cos20'sm (90°— 20°)
cos20°
2cos70°.cosec(90° -7 0 °)
+ cos(90°-20°)si n20'cos(90°— 20°)
sin20° -2cos70°.cosec(90°-70°)
= cos20°cos20° + sin20°sin20°—2*1 - 2 X 1 =1+1-2 =2-2 =0
Hence proved
(iv)We have cos80°sinio°+cos590.cosec310= 2 ^ f^ - + cos59°. cosec31° = 2sm l0 °
We will simplify the left hand side
cos80°sin10° +cos590.cosec31°-^f^ + cos59°. cosec31°=sznlO
cos(90o-ioo)sinio°+cos590.cosec(900-590) cos(s9.°l0!° + cos59°. cosec(90° — 59°)
=sinioosinioo+cos590.sec59°^i^- + cos59°.sec59° =1+1 =2s*nlO
Hence proved.
Question 7
If A,B>C are the interior of triangle ABC , show that
(i) sin(B+c2)=cosA2sin(^±^) = cos 4
Solution
A+B+C=180°
B + C= 180°-A2 4
LHS=RHS
(ii)cos(90°-A2)=sinA2cos(90° - 4 ) = sin4
LHS=RHS
Question 8
If 20+45°and3O-©20 + 45° and 30 — 0 are acute angles , find the degree measure of
0 0 satisfying sin(2O+45o)=COS(3O°+0)sin(2O + 45°) = cos(30° + 0)
Solution
Here 2O+45°=sin(6O°+0)sin(6O0 + 0)
We know that ,((9O°-0)(9O° — 0)= cos(0)cos(0)
= sin(20+45°)=sin(9Oo-(3O°-0))sin(20 + 45°) = sin(90° - (30° - 0 ))
= sin(20+45o)=sin(9Oo-3O°-0)sin(2© + 45°) = sin(90° - 30° - 0)
= sin(20+45o)=sin(6O°+0)sin(2© + 45°) = sin(60° + 0)
On equating sin of angle of we get,
= 2©+45°=6O°+©20 + 45° = 60° + 0 = 0=15°0 = 15°
Question 9
If 0 0 is appositive acute angle such that sec0=CSc6O°sec 0 = esc 60° , find
2 c o s 20 - 1 2 cos2 0 — 1
Solution
We know that, sec(9O°-0)=CSC20sec(9O° — 0) = esc2 0
= sec(©)=sec(90°-60°)sec(©) — sec(90° - 60°)
= ©=3O°0 = 30° = 2cos2©-1 2 cos2 0 - 1
= 2cos230-1 2 cos2 30 - 1
= 2(^32)2- 1 2 ( ^ ) 2 - 1
= 2(34)-12(|) - 1 =(32)-1(f) - 1 = (12)(|)
Qio.if sin30=cos(0-6°)where30and0-6circsin30 = cos(@ — 6°) where 30 and 0 — 6c*rc acute angles, find the value of 0 0 .
Soln:
We have, sin30=cos(0-6°)sin3© = cos(© — 6°)
cos(9O°+30)=cos(0-6°)cos(9O0 + 30) = cos(0 — 6°) 9O°-30=0-6°9O° 30 = 0 -6 ° -30-0=6°-9O°—30 - 0 = 6° - 90° -40=96°-4© = 96° O=-96°-4=24o0 = ^ = 24°
Q11 .If sec2A=CSC(A-42°)sec 2A = csc(A — 42°) where 2A is acute angle, find the value of A.
Soln:we know that sec(9O-3Theta)=CSC0sec (90 — 3 Theta) = esc©
sec2A=sec(90-(A-42))sec 2A = sec(90- (A - 42)) sec2A=sec(90-A+42)) sec2.A = sec(90-24 + 42)) sec2A=sec(132-A))sec2A = sec(132-.A))
Now equating both the angles we get
2A = 132 - A
A = =1323= ^
A = 44