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A Squre Study Material | Arif Baig Cell:9703806542, Email:[email protected], web:urdu4medn.webnode.com
SSC TRIGONOMETRY
SSC TR
IGO
NO
METR
Y
A SQUARE SERIES
A Square Study Material SSC Mathematics Trigonometry
2 Prepared By:ARIF BAIG Cell:9703806542, Email:[email protected]., Web: urdu4medn.webnode.com
11. Trigonometry
Triogonometry: Trigonometry is the study of relationship of three sides and three angles measures of a triangle.
Naming the sides in a right angle triangle: C Hypotenus Opposite side of angle A 𝜃 A B Adjacent side of angle A In triangle ABC ∠CAB = 𝜃(acute angle) AB= Adjacent side of angle A BC= Opposite side of angle A AC= Hypotenus C Trigonometric ratios: SinA CosecA Hypotenus Opposite side of angle A SecA TanA 𝜃 A CosA CotA B Adjacent side of angle A
Sine of ∠𝐴= SinA=𝐵𝐶
𝐴𝐶=
Opposite side to angle A
Hypotenus
Cosecant of ∠𝐴=CosecA=𝐴𝐶
𝐵𝐶=
Hypotenus
Opposite side to angle A
A Square Study Material SSC Mathematics Trigonometry
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Cosine of ∠𝐴=CosA=𝐴𝐵
𝐴𝐶=
Adjacent side to angle A
Hypotenus
Secant of ∠𝐴=SecA=𝐴𝐶
𝐴𝐵=
Hypotenus
Adjacent side to angle A
Tangent of ∠𝐴=TanA=𝐵𝐶
𝐴𝐵=
Opposite side to angle A
Adjacent side to an gle A
Cotangent of ∠𝐴=CotA=𝐴𝐵
𝐵𝐶=
Opposite side to angle A
Adjacent side to angle A
SinA and cosecA , CosA and SecA, TanA and CotA are reciprocal or multiplicative
inverse to each other.
SinA=1
𝐶𝑜𝑠𝑒𝑐𝐴 or CosecA=
1
𝑆𝑖𝑛𝐴
CosA=1
𝑆𝑒𝑐𝐴 or SecA=
1
𝐶𝑜𝑠𝐴
TanA=1
𝐶𝑜𝑡𝐴 or CotA=
1
𝑇𝑎𝑛𝐴
TanA=𝑆𝑖𝑛𝐴
𝐶𝑜𝑠𝐴 or CotA=
𝐶𝑜𝑠𝐴
𝑆𝑖𝑛𝐴
Trigonometric Ratios Of Some Specific Angles:
Degree Angle 𝐴
00 300 450 600 900
SinA 0 1
2
1
2 3
2 1
CosA 1 3
2
1
2
1
2 0
TanA 0 1
3 1 3 ∞ (Not defind)
CotA ∞ (Not defind) 3 1 1
3 0
CosecA 0 2
3 2 2 ∞ (Not defind)
SecA ∞ (Not defind) 2 2 2
3 0
Note:
The values of Sin𝜃 and Cos𝜃 always lies between 0 and 1.
A Square Study Material SSC Mathematics Trigonometry
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The values of Tan𝜃 always lies between 0 and ∞(not defind).
The values of Cot𝜃 always lies between ∞(not defind) and 0.
The values of Cosec𝜃 always lies between ∞(not defind) and 1.
The values of Sec𝜃 always lies between 1 and ∞(not defind).
Trigonometric Ratios Of Complementary Angles: If the sum of two angles is equal to 900 then the angles are called complementary angles.
In a ∆ABC if ∠𝐵=900 , then sum of other two angles must be 90o. C (∴ Sum of angles in a triangle 180o)
i,e; ∠A + ∠C = 900.
Hence∠A and ∠C are called complementary angles. If 𝜃 is acute anglethen Sin(900-A)=CosA and Cos(900-A)=SinA Tan(900-A)=CotA and Cot(900-A)=TanA Sec(900-A)=CosecA and Cosec(900-A)=SecA 900
A B
Trigonometric Identity:
Sin2A+Cos2A=1 Sec2A-Tan2A=1 Cosec2A-Cot2A=1
(SinA)2=Sin2A≠SinA2
DO THIS Identify “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles.(PAGE NO.271) 1. For angle R P Sol: in the∆PQR Opposite side=PQ
Adjacent side=QR Hypotenuse=PR Q R
2. (i) For angle X Z (ii) For angle Y Sol: in the ∆XYZ i) for angle X Opposite side=YZ X Y
Adjacent side=XZ Hypotenuse=XY iI) for angle Y Opposite side=XZ
Adjacent side=YZ Hypotenuse=XY
TRY THIS Find lengths of “Hypotenuse”, “Opposite side” and c “Adjacent side” for the given angles in the given triangles. 1. For angle C 4cm 5cm 2. For angle A Sol: By Pythagoras theorem B A
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AC2=AB2+BC2 (5)2=AB2+(4)2 25= AB2+16 AB2=25-16 AB2=9 AB2=32 AB=3cm.
DO THIS
1. Find (i) sin C (ii) cos C and C (iii) tan C in the adjacent triangle. Sol: By Pythagoras theorem 5cm 13cm AC2=AB2+BC2 (13)2=AB2+(5)2 169= AB2+25 B A AB2=169-25 AB2=144 AB2=122 AB=12cm.
i) SinC=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑐
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=
𝐴𝐵
𝐴𝐶=
12
13 , ii) CosC=
𝐴𝑑𝑗𝑎𝑐𝑒𝑛 𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑐
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=
𝐵𝐶
𝐴𝐶=
5
13 ,
iii) TanC=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑐
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑐=
𝐴𝐵
𝐵𝐶=
12
5
2. In a triangle XYZ, ∠Y is right angle, XZ = 17 m and YZ = 15 cm, then find (i) sin X (ii) cos Y (iii) tan X Sol:Given In ∆XYZ , ∠Y is right angle Z XY=17cm, YZ=15cm By Pythagoras theorem XZ2=XY2+YZ2 15cm 17cm (17)2=XY2+(15)2 289= XY2+225 XY2=289-225 Y X XY2=64 XY2=82 XY=8cm.
i) SinX=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑋
ℎ𝑦𝑝𝑜𝑡 𝑒𝑛𝑢𝑠𝑒=
𝑌𝑍
𝑋𝑍=
15
17 , ii) CosZ=
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑍
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=
𝑌𝑍
𝑋𝑍=
15
17 ,
iii) TanX=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑋
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑋=
𝑌𝑍
𝑋𝑌=
15
8
3. In a triangle PQR with right angle at Q, the value of angle P is x, PQ = 7 cm and QR = 24 cm, then find sin x and cos x.
Sol:Given In a triangle PQR with right angle at Q R angle P is x, PQ = 7 cm and QR = 24 cm By Pythagoras theorem PR2=PQ2+QR2 24cm PR2=(7)2+(24)2 PR2=49+576 X PR2=625 Q 7cm P PR2=252 PR=25cm.
Sinx=0𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑥
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=
𝑄𝑅
𝑃𝑅=
24
25 , Cosx=
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑥
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=
𝑃𝑄
𝑃𝑅=
7
25.
For angle C:
Opposite side=AB=3cm
Adjacent side=BC=4cm Hypotenuse=AC=5cm
For angle C:
Opposite side=BC=4cm
Adjacent side=AB=3cm Hypotenuse=AC=5cm
A Square Study Material SSC Mathematics Trigonometry
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TRY THESE
In a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and BC - CA = 7cm, then find sin A and tan B.
Sol:Given In a triangle ABC with right angle at C And BC+CA=23cm……(1) , BC-CA=7cm……..(2) From (1) and (2) we get BC+CA=23 B BC - CA =7 2BC=30 BC=15cm Substituting BC=15cm in (1) 15cm BC+CA=23 15+CA=23 CA=23-15=8cm. By Pythagoras theorem C A AB2=AC2+BC2 8cm AB2=(8)2+(15)2 AB2=64+225 AB2=289 AB2=172 AB=17cm.
SinA=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐴
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=
𝐵𝐶
𝐴𝐵=
15
17 , TanB=
0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐵
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐵=
𝐴𝐶
𝐵𝐶=
8
15
THINK AND DISCUSS
Discuss between your friends that
(i) sin x =4
3 for some value of angle x.
Sol: the value of sin𝜃 is always lies between o and 1. But here sin x =4
3> 1so, its does not exist.
(ii) The value of sin A and cos A is always less than 1. Why? Sol: (iii) tan A is product of tan and A. Sol: tanA is an abbreviated form of “tan of angle A”.so, tanA is not a product of tan and A. if tan separated from A has no meaning because of here A is angle.
TRY THIS
What will be the side ratios for sec A and cot A?
Sol: SecA=𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐴 , CotA=
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐴
0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐴
THINK AND DISCUSS
1.is 𝑠𝑖𝑛𝐴
𝑐𝑜𝑠𝐴 is equql to tanA?
Sol: yes, 𝑠𝑖𝑛𝐴
𝑐𝑜𝑠𝐴=tanA
∴sinA=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒, ∴cosA=
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑠𝑖𝑛𝐴
𝑐𝑜𝑠𝐴=
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 =tanA.
2.is 𝑐𝑜𝑠𝐴
𝑠𝑖𝑛𝐴 is equql to cotA?
Sol: yes, 𝑐𝑜𝑠𝐴
𝑠𝑖𝑛𝐴=cotA
∴sinA=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒, ∴cosA=
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑐𝑜𝑠𝐴
𝑠𝑖𝑛𝐴 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 =cotA
EXERCISE - 11.1 1. In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA
A Square Study Material SSC Mathematics Trigonometry
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respectively. Then, find out sin A, cos A and tan A. sol: Given the length of the sides of triangle A AB=8cm , BC=15cm and AC=17cm We have 17cm 8cm B C
sinA=𝐵𝐶
𝐴𝐶=
15
17, coasA=
𝐴𝐵
𝐴𝐶=
8
17, tanA=
𝐵𝐶
𝐴𝐵=
15
8.
15cm 2. The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25cm and RP = 24 cm respectively. Then find, tan Q -tan R. Sol: Given in ∆𝑃𝑄𝑅 PQ=7cm, QR=25cm and RP=24cm R We have
tanQ=𝑅𝑃
𝑃𝑄=
24
7 25cm
24cm
tanR= 𝑄𝑃
𝑃𝑅=
7
24 𝜃
Q 7cm P
tanQ-tanR=24
7−
7
24=
576−49
168−
527
168
3. In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = 𝜃. Then, find cos 𝜃 and tan 𝜃. Sol: Given in ∆𝐴𝐵𝐶 right angle at B. a=24units , b=25units , ∠BAC = 𝜃 A we have 𝜃 by Pythagoreans theorem 25 ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 (25)2 = 𝐴𝐵2 + (24)2 C 24 B 625=576+𝐴𝐵2 625-576=𝐴𝐵2 49=𝐴𝐵2 72=𝐴𝐵2 7cm= AB. Hence
cos 𝜃 =𝐴𝐵
𝐴𝐶=
7
25 tan 𝜃 =
𝐵𝐶
𝐴𝐵=
25
7
4. If cos A =12
13, then find sin A and tan A.
Sol: Given that C
cos A =12
13
we have 13 by Pythagoreans theorm ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 𝜃 (13)2 = (12)2 + 𝐵𝐶2 B 12 A 169=144+𝐵𝐶2 169-144=𝐵𝐶2 25=𝐵𝐶2 52 = 𝐵𝐶2 5cm=BC We have
sinA=𝐵𝐶
𝐴𝐶=
5
13 , cosA=
𝐵𝐶
𝐴𝐵=
5
12
5. If 3 tan A = 4, then find sin A and cos A. Sol: Given that 3 tan A = 4 C
tanA= 4
3
hence, 4 by Pythagoreans theorm ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 𝜃 𝐴𝐶2 = 32 + 42 B 3 A 𝐴𝐶2=9+16 𝐴𝐶2=25 𝐴𝐶2=52 AC=5cm We have
sinA=𝐵𝐶
𝐴𝐶=
4
5 , cosA=
𝐴𝐵
𝐴𝐶=
3
5.
6. If ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X. Sol: Let ∆𝐴𝐵𝐶 , ∆𝑃𝑄𝑅 X Given cosA=cosX
We have cosA=𝐴𝐵
𝐴𝐶 , cosX=
𝑋𝑌
𝑋𝑍
Then 𝐴𝐵
𝐴𝐶=
𝑋𝑌
𝑋𝑍 C A
by Pythagoreans theorm ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 and ∴ 𝑋𝑍2 = 𝑋𝑌2 + 𝑌𝑍2
𝐴𝐶 = 𝐴𝐵2 + 𝐵𝐶2 and 𝑋𝑍 = 𝑋𝑌2 + 𝑌𝑍2
So , 𝐴𝐶
𝑋𝑍=
𝐴𝐵
𝑋𝑌=
𝐵𝐶
𝑌𝑍
∵ ∠𝐴 = ∠𝑋.
7. Given cot 𝜃 =7
8, then evaluate (i)
(1+𝑠𝑖𝑛𝜃 )(1−𝑠𝑖𝑛𝜃 )
(1+𝑐𝑜𝑠𝜃 )(1−𝑐𝑜𝑠𝜃 )
(ii) 1+𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
Sol: Given cot 𝜃 =7
8 A
by Pythagoreans theorm ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 8 𝐴𝐶2 = 82 + 72 𝜃 𝐴𝐶2=64+49 B 7 C 𝐴𝐶2=113
AC= 113.
We have sin𝜃=𝐴𝐵
𝐵𝐶=
8
113 , cos𝜃=
𝐵𝐶
𝐴𝐶=
7
113
(i) (1+𝑠𝑖𝑛𝜃 )(1−𝑠𝑖𝑛𝜃 )
(1+𝑐𝑜𝑠𝜃 )(1−𝑐𝑜𝑠𝜃 )
=(1)2−sin 2𝜃
(1)2−cos 2𝜃
=(1−sin 2𝜃)
(1−cos 2𝜃) ∴ cos2𝜃=1 − sin2𝜃 ∴ sin2𝜃=1 −
cos2𝜃
=cos 2𝜃
sin 2𝜃
=cot2𝜃
=(7
8)2
=49
64.
(ii) 1+𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
A Square Study Material SSC Mathematics Trigonometry
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=1+
8
1137
113
= 113 +8
1137
113
= 113+8
113. 113
7
= 113+8
7.
8. In a right angle triangle ABC, right angle is at B, if
tan A = 3 then find the value of (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C Sol: Given that a right angle triangle ABC, right angle is at B, tan A
= 3
so B=900 , tan A = 3
tan A = tan600
A=600
We know in a triangle
A+B+C=1800
600+900+C=1800
1500+C=1800
C=1800-1500=300
(i) sin A cos C + cos A sin C
= sin600 cos300 + cos600 sin300
= 3
2 . 3
2 +
1
2.
1
2
=3
4+
1
4
=4
4= 1.
(ii) cos A cos C - sin A sin C =cos600 cos300 – sin600 sin300
= 1
2. 3
2- 3
2.
1
2
= 3
4- 3
4=0.
Exercise: 11.2
1. Evaluate the following
(i) sin450+cos450
Sol: Given
= sin450+cos450=1
2+
1
2=
1+1
2=
2
2=
2. 2
2= 2.
(ii) 𝑐𝑜𝑠 450
𝑠𝑒𝑐300+𝑐𝑜𝑠𝑒𝑐 600
Sol: Given
=𝑐𝑜𝑠 450
𝑠𝑒𝑐300+𝑐𝑜𝑠𝑒𝑐 600=
1
22
3+
2
3
=
1
22+2
3
=
1
24
3
=1
2×
3
4=
3
4 2
(iii) 𝑠𝑖𝑛300+𝑡𝑎𝑛 450−𝑐𝑜𝑠𝑒𝑐600
𝑐𝑜𝑡450+𝑐𝑜𝑠600−𝑠𝑒𝑐300
Sol: Given
=𝑠𝑖𝑛300+𝑡𝑎𝑛 450−𝑐𝑜𝑠𝑒𝑐600
𝑐𝑜𝑡 450+𝑐𝑜𝑠600−𝑠𝑒𝑐300 =
1
2+
1
1−
2
31
1+
1
2−
2
1
=1
1=1.
(iv) 2 𝑡𝑎𝑛2450+𝑐𝑜𝑠2300 − 𝑠𝑖𝑛2600
Sol: Given
=2
𝑡𝑎𝑛2450+𝑐𝑜𝑠2300 −
𝑠𝑖𝑛2600=2. (1)2+( 3
2)2 −(
3
2)2=2.(1)=2.
(v) 𝑠𝑒𝑐 2600−𝑡𝑎𝑛 2600
𝑠𝑖𝑛2300+𝑐𝑜𝑠 2300
Sol: Given
= 𝑠𝑒𝑐 2600−𝑡𝑎𝑛 2600
𝑠𝑖𝑛2300+𝑐𝑜𝑠 2300 =(2)2−( 3)2
(1
2)2+(
3
2)2
= 4−31
4+
3
4
=1
1+3
4
=14
4
=1
1=1.
2. Choose the right opinion and justify your choice.
(i) 2𝑡𝑎𝑛 300
1+𝑡𝑎𝑛 2450
(a) sin600 (b)cos600 (c) tan300 (d) sin300
Sol: Given that
= 2𝑡𝑎𝑛 300
1+𝑡𝑎𝑛 2450=2.
1
3
1+(1)2=
2
3
1+1=
2
3
2=
2
3.
1
2=
1
3
= tan300
(ii) 1−𝑡𝑎𝑛 2450
1+𝑡𝑎𝑛 2450
(a) tan900 (b)1 (c) sin450 (d) 0
Sol: Given that
=1−𝑡𝑎𝑛 2450
1+𝑡𝑎𝑛 2450=1−(1)2
1+(1)2=1−1
1+1=
0
2=0.
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(iii) 2𝑡𝑎𝑛 300
1−𝑡𝑎𝑛 2300
(a) cos600 (b)sin600 (c) tan600 (d) sin300
Sol: Given that
=2𝑡𝑎𝑛 300
1−𝑡𝑎𝑛 2300=2.
1
3
1−(1
3)2
=
2
3
1−1
3
=
2
33−1
3
=
2
32
3
=2
3.
3
2=
3
3=
3. 3
3=
3
= tan600
3. Evaluate sin600 .cos300+sin300 .cos600 . what is
the value of sin(600 + 300), what can you conclude.
Sol: Given that
= sin600 .cos300+sin300 .cos600
= 3
2. 3
2+
1
2.
1
2=
3
4+
1
4=
3+1
4=
4
4=1.
=sin(600 + 300)
= sin900=1.
∴sin600 .cos300+sin300 .cos600= sin(600 + 300)
We conclude that
∴ sinA.cosB+sinA.cosB= sin(𝐴 + 𝐵).
4.Is it right to say cos(600 + 300)=
cos600 .cos300+sin600 .sin300
Sol: Given that
cos(600 + 300)= cos600 .cos300+sin600 .sin300
cos900=1
2. 3
2−
3
2.
1
2
0= 3
4−
3
4
0=0
Yes , it is right to say cos(600 + 300)=
cos600 .cos300+sin600 .sin300 .
5. In right angle triangle PQR right angle is at Q and
PQ=6cm ∠𝑅𝑃𝑄 = 600 . Determine the length of QR
and PR.
Sol: given PQ=6cm and ∠𝑅𝑃𝑄 = 600
R
We know that in ∆𝑃𝑄𝑅
300
tan600 =𝑄𝑅
𝑃𝑄
3 =𝑄𝑅
6 Q 6cm P
6 3=QR
sin600 =𝑄𝑅
𝑃𝑅
3
2=
6 3
𝑃𝑅
PR=12cm.
6. In ∆𝑋𝑌𝑍, right angle at Y, YX=x and XZ=2x then
determine ∠𝑌𝑋𝑍 and ∠𝑌𝑍𝑋.
Sol: Given
In ∆𝑋𝑌𝑍, right angle at Y, YZ=x and XY=2x
By Pythagoreans theorem
XZ2=XY2+YZ2 Z
(2x)2=(x)2+YZ2 2x
4x2-x2=YZ2 Y x X
3x2=YZ2
3𝑥2=YZ
3𝑥=YZ
We know
In ∆𝑋𝑌𝑍
tanX=𝑌𝑍
𝑋𝑌=
3𝑥
𝑥= 3=tan600
X=600 .
tanZ=𝑋𝑌
𝑌𝑍=
𝑥
3𝑥=
1
3=tan300
Z=300.
Hence ∠𝑌𝑋𝑍=600, ∠𝑌𝑍𝑋=300.
7. Is it right to say that sin(A+B)=sinA+sinB justify
your answer.
Sol: Given sin(A+B)=sinA+sinB
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Let A=300 , B=600
Sin(300 + 600) = 𝑠𝑖𝑛300 + 𝑠𝑖𝑛600
sin900=1
2+
3
2
1= 1+ 3
2
1≠1+ 3
2
Isn’t right to say sin(A+B)=sinA+sinB.
Exercise: 11.3
1. Evaluate
(i) 𝑡𝑎𝑛 360
𝑐𝑜𝑡540
Sol: Given that
=𝑡𝑎𝑛 360
𝑐𝑜𝑡540
∴ cotA=tan(900 −A)
∵ cot360=tan(900 − 540)=tan360
= 𝑡𝑎𝑛 360
𝑐𝑜𝑡 360
=1.
(ii) cos120 − 𝑠𝑖𝑛780
Sol: Given that
= cos120 − 𝑠𝑖𝑛780
∴ cosA=sin(900 −A)
∵ cos120=sin(900 − 120)=sin780
= sin780 − 𝑠𝑖𝑛780
=1.
(iii) cos𝑒𝑐310 − 𝑠𝑒𝑐590
Sol: Given that
= cos𝑒𝑐310 − 𝑠𝑒𝑐590
∴ cosecA=sec(900 −A)
∵ cos𝑒𝑐310=sec(900 − 310)=sec590
= s𝑒𝑐590 − 𝑠𝑒𝑐590
=1.
(iv) sin150 . 𝑠𝑒𝑐750
Sol: Given that
= sin150 . 𝑠𝑒𝑐750
= 𝑠𝑖𝑛150
𝑐𝑜𝑠750
∴ cosA=sin(900 −A)
∵ cos750=sin(900 − 750)=sin150
= 𝑠𝑖𝑛150
𝑠𝑖𝑛150
=1.
(v) tan260 . tan640
Sol: Given that
= tan260 . tan640
∴ tanA=cot(900 −A)
∵ tan640=cot(900 − 640)=cot260
= tan260 . cot260
= tan260 .1
tan 260
=1.
2. (i) Show that tan480 . tan160 . tan420 . tan740=1
Sol: LHS. = tan480 . tan160 . tan420 . tan740
∴ tanA=cot(900 −A)
∵ tan420=cot(900 − 420)=cot480
∵ tan740=cot(900 − 740)=cot160
= tan480 . tan160 . cot480 . cot160
= tan480 . tan160 . 1
tan 480 .1
tan 160
=1. RHS
(ii) cos360 .cos540-sin360 .sin540=0
Sol: LHS
= cos360 .cos540 −sin360 .sin540
∴ cosA=sin(900 −A)
∴ cotA=tan(900 −A)
∵ cot360=tan(900 − 540)=tan360
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∵ cos360=sin(900 − 360)=sin540
∵ cos540=sin(900 − 540)=sin360
= sin540 .sin360 − sin360 . sin540
=1. RHS
3. If tan2A=cot(𝐴 − 180), where 2A is an acute
angle. Find the value of ‘A’.
Sol: Given that
=tan2A=cot(𝐴 − 180)
∴ tanA=cot(900 −A)
∵tan2A=cot(900 − 2A)
cot(900 − 2A)= cot(𝐴 − 180)
900 − 2A= 𝐴 − 180
900+180=A+2A
1080=3A
360=A.
4. If tanA = cotB where A and B are acute angles
prove that A+B=900
Sol: Given that
tanA = cotB
∴ cotA=tan(900 −A)
∵ cotB=tan(900 − 𝐵)
tanA= tan(900 − 𝐵)
A=900 − 𝐵
A+B=900 .
5. If A,B and C are the interior angles of a triangle
ABC then show that tan(𝐴+𝐵
2) = 𝑐𝑜𝑡
𝐶
2 .
Sol: We know that in triangle ABC
A+B+C= 1800 (dividing the equation by ‘2’)
𝐴+𝐵+𝐶
2=
1800
2
𝐴+𝐵
2+
𝐶
2=900
𝐴 + 𝐵
2= 900 −
𝐶
2
LHS. =tan(𝐴+𝐵
2) ∴ tanA=cot(900 −A)=cotA
=tan(900 −𝐶
2)
= 𝑐𝑜𝑡𝐶
2 RHS.
6. Express sin750 +cos650 in terms of
trigonometric ratios of angle between 00 and 450 .
Sol: Given that
= sin750 +cos650
∴ sinA=cos(900 −A)
∵ cos750=sin(900 − 750)=sin150
∴ cosA=sin(900 −A)
∵ cos650=sin(900 − 650)=sin250
= cos150 +sin250 .
Exercise: 11.4
1.Evaluate the following
(i) (1+tan 𝜃 -sec 𝜃)(1+cot 𝜃 -cosec 𝜃)
Sol: Given that (1+tan 𝜃 -sec 𝜃)(1+cot 𝜃 -
cosec 𝜃)
=(1+𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃−
1
𝑐𝑜𝑠𝜃 )(1+
𝑐𝑜𝑠𝜃
𝑠𝑖𝑛𝜃−
1
𝑠𝑖𝑛𝜃 )
= 𝑐𝑜𝑠𝜃+𝑠𝑖𝑛𝜃−1
𝑐𝑜𝑠𝜃
𝑠𝑖𝑛𝜃+𝑐𝑜𝑠𝜃−1
𝑠𝑖𝑛𝜃
= (𝑐𝑜𝑠𝜃+sin θ)2− (1)2
𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃
= 𝑐𝑜𝑠 2𝜃+𝑠𝑖𝑛2𝜃+2𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃−1
𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃 ∴ 𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠2𝜃=1
= ∤+2𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃−∤
𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃
=2𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃
=2.
(𝑖𝑖) (𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃)2 + (𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃)2
𝑠𝑜𝑙: (𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃)2 + (𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃)2
=𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠2𝜃+2sin 𝜃.cos 𝜃 +𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠2 𝜃 -2sin 𝜃.cos 𝜃
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=1+1
=2.
(iii) ( 𝑠𝑒𝑐2𝜃 − 1)( 𝑐𝑜𝑠𝑒𝑐2𝜃-1)
Sol: ( 𝑠𝑒𝑐2𝜃 − 1)( 𝑐𝑜𝑠𝑒𝑐2𝜃-1)
∴ 𝑡𝑎𝑛2= 𝑠𝑒𝑐2-1 ∴ 𝑐𝑜𝑡2=𝑐𝑜𝑠𝑒𝑐2-1
=𝑡𝑎𝑛2𝜃. 𝑐𝑜𝑡2 𝜃
=𝑡𝑎𝑛2. 1
𝑡𝑎𝑛 2
=1.
2. Show that ( 𝑐𝑜𝑠𝑒𝑐𝜃 − cot𝜃)2= 1−𝑐𝑜𝑠𝜃
1+𝑐𝑜𝑠𝜃
Sol: LHS.
=( 𝑐𝑜𝑠𝑒𝑐𝜃 − cot𝜃)2
= 1
𝑠𝑖𝑛𝜃−
𝑐𝑜𝑠𝜃
𝑠𝑖𝑛𝜃
2
= 1−𝑐𝑜𝑠𝜃
𝑠𝑖𝑛𝜃
2
= (1−𝑐𝑜𝑠𝜃)2
𝑠𝑖𝑛2 𝜃
= (1−𝑐𝑜𝑠𝜃)2
12−𝑐𝑜𝑠 2𝜃
= (1−cos 𝜃)(1−cos 𝜃)
(1−cos 𝜃)(1+cos 𝜃)
= (1−cos 𝜃)
(1+cos 𝜃). RHS
3.Show that 1+𝑠𝑖𝑛𝐴
1−𝑠𝑖𝑛𝐴 =secA+tanA
Sol: LHS. 1+𝑠𝑖𝑛𝐴
1−𝑠𝑖𝑛𝐴
Multiply and divided by (1+sinA)
= 1+𝑠𝑖𝑛𝐴
1−𝑠𝑖𝑛𝐴×
1+𝑠𝑖𝑛𝐴
1+𝑠𝑖𝑛𝐴
= (1+𝑠𝑖𝑛𝐴 )2
12−𝑠𝑖𝑛𝐴2
= (1+𝑠𝑖𝑛𝐴 )2
𝑐𝑜𝑠𝐴2
=1+𝑠𝑖𝑛𝐴
𝑐𝑜𝑠𝐴
=1
𝑐𝑜𝑠𝐴+
𝑠𝑖𝑛𝐴
𝑐𝑜𝑠𝐴 ∴ 𝑠𝑒𝑐𝐴 =
1
𝑐𝑜𝑠𝐴 ∴tanA=
𝑠𝑖𝑛𝐴
𝑐𝑜𝑠𝐴
=secA+tanA. RHS
4.Show that 1−𝑡𝑎𝑛 2𝐴
𝑐𝑜𝑡 2𝐴−1= 𝑡𝑎𝑛2𝐴.
Sol: LHS. = 1−𝑡𝑎𝑛 2𝐴
𝑐𝑜𝑡 2𝐴−1
=1−𝑡𝑎𝑛 2𝐴
1
𝑡𝑎𝑛 2𝐴−
1
1
=1−𝑡𝑎𝑛 2𝐴
1
1−𝑡𝑎𝑛 2𝐴
𝑡𝑎𝑛 2𝐴
= 1−𝑡𝑎𝑛 2𝐴
1×
𝑡𝑎𝑛 2𝐴
1−𝑡𝑎𝑛 2𝐴
= 𝑡𝑎𝑛2𝐴. RHS.
5. Show that 1
𝑐𝑜𝑠𝜃− 𝑐𝑜𝑠𝜃= tan 𝜃.sin 𝜃
Sol: LHS. = 1
𝑐𝑜𝑠𝜃− 𝑐𝑜𝑠𝜃
= 1−𝑐𝑜𝑠 2𝜃
𝑐𝑜𝑠𝜃 ∴ 𝑠𝑖𝑛2𝜃 = 1 − 𝑐𝑜𝑠2𝜃
= 𝑠𝑖𝑛2𝜃
𝑐𝑜𝑠𝜃
= 𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃 .sin 𝜃
= tan 𝜃.sin 𝜃 RHS.
6. Simplify secA(1-sinA)(secA+tanA).
Sol: secA(1-sinA)(secA+tanA)
= (secA-secA.sinA)(secA+tanA)
= (secA-𝑠𝑖𝑛𝐴
𝑐𝑜𝑠𝐴)(secA+tanA)
= (secA-tanA)(secA+tanA)
= 𝑠𝑒𝑐2𝐴 − 𝑡𝑎𝑛2𝐴 ∴ 𝑠𝑒𝑐2𝐴 − 𝑡𝑎𝑛2𝐴 =1
= 1.
7. Prove that (𝑠𝑖𝑛𝐴 + 𝑐𝑜𝑠𝑒𝑐𝐴)2 + (𝑐𝑜𝑠𝐴 +
𝑠𝑒𝑐𝐴)2=7+𝑡𝑎𝑛2𝐴+𝑐𝑜𝑡2𝐴.
Sol: LHS. = (𝑠𝑖𝑛𝐴 + 𝑐𝑜𝑠𝑒𝑐𝐴)2 + (𝑐𝑜𝑠𝐴 + 𝑠𝑒𝑐𝐴)2
=𝑠𝑖𝑛2𝐴 + 𝑐𝑜𝑠𝑒𝑐𝐴2+2.sinA.cosecA+𝑐𝑜𝑠2𝐴 +
𝑠𝑒𝑐𝐴2+2.cosA.secA
= 𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴+𝑐𝑜𝑠𝑒𝑐𝐴2+𝑠𝑒𝑐𝐴2+2+2
= 1+1+ 𝑐𝑜𝑡𝐴2+1 + 𝑡𝑎𝑛𝐴2+4 ∴ 𝑐𝑜𝑠𝑒𝑐𝐴2
=1+ 𝑐𝑜𝑡𝐴2 ∴ 𝑠𝑒𝑐𝐴2 = 1 + 𝑡𝑎𝑛𝐴2
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= 7+𝑡𝑎𝑛𝐴2 + 𝑐𝑜𝑡𝐴2. RHS.
8. Simplify (1-cos 𝜃)(1+cos 𝜃)(1+𝑐𝑜𝑡2𝜃)
Sol: Given that
=(1-cos 𝜃)(1+cos 𝜃)(1+𝑐𝑜𝑡2𝜃)
=(12 − 𝑐𝑜𝑠2𝜃)(1 + 𝑐𝑜𝑡2𝜃)
=𝑠𝑖𝑛2𝜃(1+𝑐𝑜𝑠 2𝜃
𝑠𝑖𝑛2𝜃)
=𝑠𝑖𝑛2𝜃(𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠 2𝜃
𝑠𝑖𝑛2𝜃) ∴ 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃=1
= 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃
=1.
9. sec 𝜃 +tan 𝜃 =p then what is the value of sec 𝜃 -
tan 𝜃 ?
Sol: Given that
sec 𝜃 +tan 𝜃 =p
∴ sec 𝜃 -tan 𝜃 =1
sec 𝜃+tan𝜃
= sec 𝜃 -tan 𝜃 =1
p
10. If cosec 𝜃 +cot 𝜃 =k then prove that cos 𝜃 =𝑘2−1
k2+1
Sol: Given that
cosec 𝜃 +cot 𝜃 =k
∴ 𝑐𝑜sec 𝜃 -cot 𝜃 =1
cosec 𝜃+cot 𝜃
= cosec 𝜃 -cot 𝜃 =1
k
From and
cosec 𝜃 +cot 𝜃 =k
cosec 𝜃 -cot 𝜃 =1
k
2cosec 𝜃 = k+1
k
2cosec 𝜃 = 𝑘2+1
𝑘
cosec 𝜃 = 𝑘2+1
2𝑘
∴ 𝑐𝑜𝑠2𝜃 =1- 𝑠𝑖𝑛2𝜃
cos 𝜃 = 1 − 𝑠𝑖𝑛2𝜃
cos 𝜃 = 1 − 2k
k2+1
2
cos 𝜃 = (k2+1)2 −4k2
(k2+1)2
cos 𝜃 = (k2−1)2
(k2+1)2
cos 𝜃 = (k2−1)2
(k2+1)2 RHS.
1
1
1
2
2
2
1
1
1