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Trigonometry
1.TRIGONOMETRIC RATIOS
→
1. If a ray OP makes an angle θ with the positive direction of X-axis then
y
x
ii) cos θ =
i) Sin θ =
r
r
x
y
iii) tan θ = (x ≠ 0)
iv) cot θ =
(y ≠ 0)
y
x
y
r
r
(x ≠ 0)
vi) cosec θ =
(y ≠ 0)
v) sec θ =
x
y
2. Relations :
i) sin θ cosec θ = 1
ii) cos θ sec θ = 1
O
iii) tan θ cot θ = 1
iv) sin2 θ + cos2 θ = 1
→ (sec θ + tan θ) (sec θ – tan θ) = 1.
v) 1 + tan2 θ = sec2 θ
1
→ sec θ + tan θ =
=1
sec θ − tan θ
vi) 1 + cot2 θ = cosec2 θ → (cosec θ + cot θ) (cosec θ – cot θ) = 1
1
→ cosec θ + cot θ =
cosec θ − cot θ
2
2
vii) sec θ + cosec θ = sec2 θ . cosec2 θ
viii) tan2 θ – sin2 θ = tan2 θ . sin2 θ;
cot2 θ – cos2 θ = cot2 θ . cos2 θ
ix) sin2 θ + cos4 θ = 1 – sin2 θ cos2 θ
= sin4 θ + cos2 θ
x) sin4 θ + cos4 θ = 1 – 2sin2 θ cos2 θ
xi) sin6 θ + cos6 θ = 1 – 3sin2 θ cos2 θ
xii) sin2 x + cosec2 x ≥ 2
xiii) cos2 x + sec2 x ≥ 2
xiv) tan2 x + cot2 x ≥ 2.
3. Values of trigonometric ratios of certain angles
↓
ratio
angle
→
sin
cos
0o
0
1
π/6
π/4
1/2 1/ 2
3/2 1/ 2
tan
0
1/ 3
cot
undefined
3
cosec undefined
2
sec
1
2/ 3
1
1
2
2
π/3
3/2
1/2
π/2
1
0
3 undefined
1/ 3
0
2/ 3
1
2
undefined
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P
r
y
θ
x
x
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4. Signs of Trigonometric ratios : If θ lies in I, II, III, IV quadrants then the signs of
trigonometric ratios are as follows.
II
90o < θ < 180o
sin θ and cosec θ
are (+)ve
I
0o < θ < 90o
all the ratios
are (+)ve
III
180o < θ < 270o
tan θ and cot θ
are (+)ve
IV
270o < θ < 360o
cos θ and sec θ
are (+)ve
Note : i) 0o, 90o, 180o, 270o. 360o, 450o, ….. etc. are called quadrant angles.
ii) With “ALL SILVER TEA CUPS” symbol we can remember the signs of
trigonometric ratios.
5. Coterminal angles : If two angles differ by an integral multiples of 360o then two angles
are called coterminal angles.
Thus 30o, 390o, 750o, 330o etc., are coterminal angles.
Fn
sin θ
cos θ
90 ∓ θ
cos θ
± sin θ
± cot θ
tan θ
cosec θ
sec θ
± cosec θ
sec θ
cot θ
± tan θ
180 ∓ θ
± sin θ
− cos θ
270 ∓ θ
− cos θ
∓ sin θ
360 ∓ θ
∓ sin θ
cos θ
∓ tan θ
∓ tan θ
± cot θ
± cosec θ − sec θ ∓ cosec θ
− sec θ ∓ cosec θ
sec θ
∓ cot θ
± tan θ
∓ cot θ
6. Complementary Angles : Two Angles A, B are said to complementary ⇒ A + B = 90o
7. Supplementary angles : Two angles A, B are said to be supplementary
⇒ A + B = 180o.
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PROBLEMS
VSAQ’S
Convert the following into simplest form :
1. tan(θ – 14π)
Sol. –tan(14π – θ)
= tan[7 ⋅ 2π – θ] = –[–tan θ] = tan θ
2.
cosec (5π + θ)
Sol. cosec (5π + θ) = –cosec θ
(∵ cos ec(nπ + θ) = (−1) n cos ecθ)
⎛ 7π ⎞
Find the value of cos ⎜ − ⎟
⎝ 2 ⎠
7π
⎛ 7π ⎞
Sol. cos ⎜ − ⎟ = cos
=0
2
⎝ 2 ⎠
3.
(∵ cos(2n + 1)
π
= 0)
2
4. Find the value of cot(315°)
Sol. cot(315°) = – cot 315°
= –[cot 360° – 45°]
= – [– cot 45°]
= cot 45° = 1
5.
Evaluate
cos245° + cos2135° + cos2 225° + cos2 315°
1
Sol. cos 45° =
, cos135° = cos(180° − 45°)
2
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= cos 45°
=−
1
2
cos 225° = cos(180° + 45°)
= cos 45°
=−
1
2
cos135° = cos(360° − 45°)
1
= cos 45°
=
2
Given expression
2
2
2
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
⎜
⎟ +⎜−
⎟ +⎜−
⎟ +⎜
⎟
2⎠ ⎝
2⎠ ⎝ 2⎠
⎝ 2⎠ ⎝
1 1 1 1
= + + + =2
2 2 2 2
2
6. cos 225° – sin 225° + tan 495° – cot 495°
Sol. cos 225° = cos(180° + 45°)
1
= cos 45° = −
2
sin 225° = sin(180° + 45°)
−1
2
tan 495° = tan[5(90°) + 45]
= − sin 45° =
= − cot 45° = −1
cot 495° = cot[5(90°) + 45°]
= − tan 45° = −1
G.E.
−1
2
−1
=
2
=
⎛ 1 ⎞
−⎜−
⎟ −1+1
2⎠
⎝
1
+
−1 +1 = 0
2
If cos θ = t (0 < t < 1) and θ does not lie in the first quadrant, find the value of (a) sinθ
(b) tan θ.
Sol. cos θ = t ⇒ lies in IV quadrant.
7.
x 2 = AC2 − BC2 = 1 − t 2
x = 1− t2
a) sin θ =
AB
= − 1− t2
AC
AB
1− t2
=−
b) tan θ =
BC
t
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8.
1
and θ does not lie in third quadrant, find the values of
3
(i) cos θ (ii) cot θ
If sin θ = −
Solution:
1
θ does not lie in III Q
3
∴θ lies in IV Quadrant
sin θ = −
cos θ = 1 − sin 2 θ = 1 −
1 2 2
=
9
3
cos θ
=2 2
sin θ
9. Find the value of sin 3300 . cos1200 + cos 2100 . sin 3000
Solution:
1
sin 3300 = sin ( 3600 − 300 ) = − sin 300 = −
2
1
3
3
cos1200 = − cos 2100 = −
; sin 3000 = sin ( 3600 − 600 ) = −
2
2
2
1
1 ⎛
3⎞⎛
3⎞
sin 3300 cos1200 + cos 2100 . sin 3000 = − × − + ⎜⎜ −
⎟⎟ ⎜⎜ −
⎟ =1
2
2 ⎝ 2 ⎠ ⎝ 2 ⎟⎠
1
10. If cosec θ + cot θ = , find cos θ and determine the quadrant in which θ lies.
3
2
2
Sol. cosec θ – cot θ = 1
(cosec θ + cot θ) (cosec θ – cot θ) = 1
cosec θ – cot θ = 3
…(1)
1
…(2)
and cosec θ + cot θ =
3
From (1) + (2)
cosec θ – cot θ = 3
1
cosec θ + cot θ =
3
1 10
2cosec θ = 3 + =
3 3
10
cos ecθ =
6
6
sin θ =
16
cot θ =
From (1) – (2)
cosec θ – cot θ = 3
cosec θ + cot θ =
1
3
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1 8
−2 cot θ = 3 − =
3 3
−8
8
=
cot θ = −
2×3 6
∴ cos θ = cot θ ⋅ sin θ =
−8 6 −4
× =
6 10 5
Thus sin θ is +ve and cos θ is –ve
Then θ lies in II quadrant.
11. If sec θ + tan θ = 5, find the quadrant in which θ lies and find the value of sin θ.
Sol. sec θ + tan θ = 5
…(1)
We know, sec2 θ – tan2 θ = 1
(sec θ + tan θ)(sec θ – tan θ) = 1
1
sec θ + tan θ
1
sec θ − tan θ =
...(2)
5
Adding (1), (2)
sec θ − tan θ =
sec θ + tan θ = 5
sec θ − tan θ =
1
5
1
5
25 + 1
2sec θ =
5
26 1
sec θ = ×
5 2
13
sec θ =
5
12
12
tan θ = , sin θ =
5
13
2sec θ = 5 +
∴ sin θ ⋅ sec θ and tan θ are positives
∴ θ lies in I quadrant.
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1
12. If cos A = cos B = − : A does not lie in second quadrant B does not lie in third quadrant
2
4sin B − 4 tan A
find the value of
tan B + sin A
Solution:
1
cos A = cos B = − ;
A does not lie in II Q
2
∴ A lies in III Q
3
sin A = −
2
1
cos B = −
B does not lie in III Q
2
∴ B lies in II Q
sin B =
3
2
tan A = 3
4sin B − 3 tan A
=
tan B + sin A
13.
tan B = − 3
3
− 3× 3
3
2
2
=−
=
−3 3 3
3
− 3−
2
2
4×
π⎞
⎛
⎛ 3π
⎞
sin(3π − A) cos ⎜ A − ⎟ tan ⎜ − A ⎟
2⎠
⎝
⎝ 2
⎠ = − cos 4 A
13
π
π⎞
⎛
⎞
⎛
cos ec ⎜
+ A ⎟ sec(3π + A) cot ⎜ A − ⎟
2⎠
⎝ 2
⎠
⎝
Sol. sin(3π – A) = sin A
π⎞
⎛
⎛π
⎞
cos ⎜ A − ⎟ = cos ⎜ − A ⎟ = sin A
2⎠
⎝
⎝2
⎠
⎛ 3π
⎞
tan ⎜ − A ⎟ = cot A
⎝ 2
⎠
⎛ 13π
⎞
cos ec ⎜
+ A ⎟ = − sec A
⎝ 2
⎠
sec(3π + A) = − sec A
π⎞
⎛
⎛π
⎞
cot ⎜ A − ⎟ = − cot ⎜ − A ⎟ = − tan A
2⎠
⎝
⎝2
⎠
sin A ⋅ sin A ⋅ cot A
L.H.S. =
− sec A ⋅ − sec A ⋅ − tan A
cos A
sin 2 A ×
sin A
=
1
sin A
−
×
2
cos A cos A
cos3 A
= sin A cos A ×
= − cos 4 A
− sin A
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14. cot ⎛⎜ π ⎞⎟ cot ⎛⎜ 3π ⎞⎟ cot ⎛⎜ 5π ⎞⎟ cot ⎛⎜ 7 π ⎞⎟ cot ⎛⎜ 9π ⎞⎟ = 1
⎝ 20 ⎠ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎝ 20 ⎠
1
⎛ π ⎞
Sol. cot ⎜ ⎟ = cot 9° =
tan 9°
⎝ 20 ⎠
1
⎛ 3π ⎞
cot ⎜ ⎟ = cot 27° =
tan 27°
⎝ 20 ⎠
⎛ 5π ⎞
cot ⎜ ⎟ = cot 45° = 1
⎝ 20 ⎠
⎛ 7π ⎞
cot ⎜ ⎟ = cot 63° = cot(90° − 27°) = tan 27°
⎝ 20 ⎠
⎛ 9π ⎞
cot ⎜ ⎟ = cot 81° = cot(90° − 9°) = tan 9°
⎝ 20 ⎠
π
3π
5π
7π
9π
cot cot cot
cot
20
20
20
20
20
1
1
=
⋅1⋅ tan 27°⋅ tan 9° = 1
tan 9° tan 27°
∴ cot
⎛ 11π ⎞
⎛ 35π ⎞ ⎛ 7 π ⎞
sin ⎜ −
⎟ tan ⎜
⎟ sec ⎜ − ⎟
3 ⎠
6 ⎠ ⎝ 3 ⎠
⎝
⎝
15. Simplify
.
⎛ 5π ⎞
⎛ 7π ⎞
⎛ 17π ⎞
cos ⎜ ⎟ cos ec ⎜ ⎟ cos ⎜
⎟
⎝ 4 ⎠
⎝ 4 ⎠
⎝ 6 ⎠
11π
⎛ 11π ⎞
Sol. sin ⎜ −
= − sin 660°
⎟ = − sin
3
⎝ 3 ⎠
= − sin(2 ⋅ 360° − 60°)
= sin 60° =
3
2
1
⎛ 35π ⎞
tan ⎜
⎟ = tan1050° = tan(3 ⋅ 360° − 30°) = − tan 30° = −
3
⎝ 6 ⎠
⎛ 7π ⎞
sec ⎜ − ⎟ = sec(−420°) = sec 420° = sec(360° + 60°) = sec 60° = 2
⎝ 3 ⎠
⎛ 5π ⎞
cos ⎜ ⎟
⎝ 4 ⎠
= cos(225°) = cos(180° + 45°) = cos 45° = −
⎛ 7π ⎞
csc ⎜ ⎟
⎝ 4 ⎠
= csc 315° = csc(360° − 45°) = − csc 45° = − 2
1
2
⎛ 17 π ⎞
cos ⎜
⎟ = cos 510° = cos(360° + 150°)
⎝ 6 ⎠
= cos100° = cos(180° − 30°) = − cos 30° = −
3
2
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⎛ 3 ⎞ ⎛ −1 ⎞
⎜
⎟⎜
⎟ (2)
2
3
⎝
⎠
⎠
L.H.S. = ⎝
⎛
3⎞
⎛ −1 ⎞
(
2)
−
−
⎜
⎟
⎜
⎟
⎝ 2⎠
⎝ 2 ⎠
=
2
−1
=
= R.H.S.
3
− 3
2
tan 610° + tan 700° 1 − p 2
=
16. If tan 20° = p, prove that
.
tan 560° − tan 470° 1 + p 2
Sol.
tan 610° + tan 700°
tan 560° − tan 470°
tan(360° + 250°) + tan(360° + 340°)
=
tan(360° + 200°) − tan(360° + 110°)
tan 250° − tan 340°
tan 200° − tan110°
tan(270° − 20°) − tan(360° − 20°)
=
tan(180° + 20°) + tan(90° + 20°)
=
cot 20° − tan 20°
tan 20° + cot 20°
1
− tan 20°
°
tan
20
=
1
tan 20° +
tan 20°
=
1
1 − p2
−p
1 − p2
p
p
=
=
=
1 p2 + 1 p2 + 1
p+
p
p
∴
tan 610° + tan 700° 1 − p 2
=
tan 560° − tan 470° 1 + p 2
17. If α, β are complementary angles such that b sin α = a, then find the value of
(sin α cos β – cos α sin β).
Sol. α and β are complementary angles.
α + β = 90°
β = 90° – α
b sin α = a ⇒ sin α =
a
b
2
b2 − a 2
⎛a⎞
cos α = 1 – sin α = 1 − ⎜ ⎟ =
b2
⎝b⎠
2
2
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b2 − a 2
cos α =
b
Since β = 90° – α
sin β = sin(90 − α) = cos α
∴ cos α = sin β
and α = 90° − β
sin α = sin(90° − β) = cos β
∴ cos β = sin α =
a
b
a
b
sin α cos β − cos α sin β
∴ cos β =
a a
b2 − a 2
b2 − a 2
= × −
×
b b
b
b
a 2 (b 2 − a 2 ) a 2 − b 2 + a 2 2a 2 − b 2
= 2−
=
=
b
b2
b2
b2
18. If cos θ + sin θ = 2 cos θ then prove that cos θ − sin θ = 2 sin θ
Solution:
cos θ + sin θ = 2 cos θ -------- (1) squaring and adding (1) & (2)
Let cos θ − sin θ = x -------(2)
( cos θ + sin θ )
2
+ ( cos θ − sin θ ) =
2
(
2 cos θ
⇒ 2 − 2 cos 2 θ = x 2 ⇒ x = 2 sin θ
)
2
+ x2
∴ cos θ − sin θ = 2 sin θ
19. If 8 tan A = 15 and 25 sin B = –7 and neither A nor B is in the fourth quadrant, then show
304
that sin A cos B + cos A sin B = −
.
425
Sol. 8 tan A = –15
15
tan A = −
8
A lies in II quadrant
15
−8
sin A = , cos A =
17
17
25sin B = −7
7
25
B lies in III quadrant
7
−24
sin B = − , cos B =
25
25
L.H.S. : sin A cos B + cos A sin B
sin B = −
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15 −24 −8 −7 −3 × 24 −8 × −7
×
+ ×
=
+
17 25 17 25 17 × 5 17 × 25
−360 56 −304
=
+
=
425 425 425
π
4π
6π
9π
20. Prove that sin 2
+ sin 2
+ sin 2
+ sin 2
=2
10
10
10
10
Solution:
π
4π
6π
9π
sin 2
+ sin 2
+ sin 2
+ sin 2
10
10
10
10
π
π ⎞
⎛π π ⎞
⎛π π ⎞
⎛
sin 2 + sin 2 ⎜ − ⎟ + sin 2 ⎜ + ⎟ + sin 2 ⎜ π − ⎟
10
10 ⎠
⎝ 2 10 ⎠
⎝ 2 10 ⎠
⎝
π
π
π
π
π
π⎫
⎧
sin 2 + cos 2
+ cos 2 + sin 2 = 2 ⎨sin 2 + cos 2 ⎬ = 2
10
10
10
10
10
10 ⎭
⎩
=
= sin 2
21. If tan 200 = λ then show that
22. If sec θ + tan θ =
π
10
+ cos 2
π
10
=1
tan 1600 − tan1100 1 − λ 2
=
1 + tan 1600 tan 1100
2λ
2
find the value of sin θ and determine the quadrant in which θ lies
3
Solution:
2
3
2 3
2sec θ = + ⇒ sec θ = 13 /12
3 2
∴θ lies in IV Quadrant
sec θ + tan =
3
2
tan θ −15
=
sin θ =
sec θ
13
sec θ + tan θ =
23. If A, B, C, D are angles of a cyclic quadrilateral then prove that
i) sin A – sin C = sin D – sin B
ii) cos A + cos B + cos C + cos D = 0
Sol. A, B, C, D are the angles of a cyclic quadrilateral.
A + C = 180°; B + D = 180°
A = 180° – C, B = 180° – D
i) sin A – sin C = sin D – sin B
sin A + sin B = sin C + sin D
we have
A = 180° – C ⇒ Sin A = sin C …(1)
B = 180° – D ⇒ sin B = sin D …(2)
Adding (1) and (2)
Sin A + sin B = sin C + sin D
ii) cos A = cos (180° – C) = – cos C
⇒ cos A + cos C = 0
…(1)
cos B = cos (180° – D) = – cos D
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⇒ cos B + cos D = 0
…(2)
Adding (1) and (2)
cos A + cos B + cos C + cos D = 0.
24. If a cos θ – b sin θ = c, then show that a sin θ + b cos θ = ± a 2 + b 2 − c 2 .
…(1)
Sol. a cos θ – b sin θ = c
Let a sin θ + b cos θ = k …(2)
Squaring and adding
25. If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ±3.
Sol. 3 sin A + 5 cos A = 5
Let 5 sin A – 3 cos A = k
Squaring and adding
(3 sin A + 5 cos A)2 + (5 sin A – 3 cos A)2 = 25 + k2
9sin 2 A + 25cos 2 A + 30sin A cos A + 25sin 2 A + 9 cos 2 A − 30sin A cos A = 25 + k 2
34sin 2 A + 34 cos 2 A = 25 + k 2
34(sin 2 A + cos 2 A) = 25 + k 2
34(1) = 25 + k 2
34 = 25 + k 2
k 2 = 34 − 26 = 9
k = ±3
tan θ + sec θ − 1 1 + sin θ
=
.
tan θ − sec θ + 1
cos θ
tan θ + sec θ − 1
Sol. We have
tan θ − sec θ + 1
26. Prove that
(tan θ + sec θ) − (sec 2 θ − tan 2 θ)
tan θ − sec θ + 1
(tan θ + sec θ) − (sec θ + tan θ)(sec θ − tan θ)
=
tan θ − sec θ + 1
[tan θ + sec θ][1 − sec θ + tan θ]
=
tan θ − sec θ + 1
= tan θ + sec θ
=
sin θ
1
1 + sin θ
+
=
cos θ cos θ
cos θ
tan θ + sec θ − 1 1 + sin θ
∴
=
tan θ − sec θ + 1
cos θ
=
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27. Prove that (1 + cot θ − csc θ)(1 + tan θ + sec θ) = 2 .
Sol. L.H.S. : (1 + cot θ − csc θ)(1 + tan θ + sec θ)
1 ⎤ ⎡ sin θ
1 ⎤
⎡ cos θ
= ⎢1 +
−
1+
+
⎥
⎢
⎣ sin θ sin θ ⎦ ⎣ cosθ cos θ ⎥⎦
⎡ sin θ + cos θ − 1 ⎤ ⎡ cos θ + sin θ + 1 ⎤
=⎢
⎥⎦ ⎢⎣
⎥⎦
sin θ
cos θ
⎣
(sin θ + cos θ) 2 − 12 sin 2 θ + cos 2 θ + 2sin θ cos θ − 1
=
sin θ cos θ
sin θ cos θ
1 + 2sin θ cos θ − 1
sin θ cos θ
=
=2
= 2 = R.H.S.
sin θ cos θ
sin θ cos θ
Eliminate θ from the following equations.
=
28. x = a cos3θ ; y = b sin3 θ
Sol. x = a cos3θ ; y = b sin3 θ
1/ 3
x
⎛x⎞
= cos3 θ ⇒ cos θ = ⎜ ⎟
a
⎝a⎠
1/ 3
y
⎛y⎞
= sin 3 θ ⇒ sin θ = ⎜ ⎟
b
⎝b⎠
We have cos2 θ + sin2 θ = 1
2/3
2/3
⎛x⎞
⎛y⎞
⇒ ⎜ ⎟ +⎜ ⎟ =1
⎝a⎠
⎝b⎠
29. x = a cos4 θ = b sin4 θ.
Sol. x = a cos 4 θ ⇒ cos 4 θ =
⇒ cos 2 θ =
x
a
y = b sin 4 θ ⇒ sin 4 θ =
⇒ sin 2 θ =
x
a
y
b
y
b
cos 2 θ + sin 2 θ = 1
1/ 2
x
y
⎛x⎞
+
=1⇒ ⎜ ⎟
a
b
⎝a⎠
1/ 2
⎛y⎞
+⎜ ⎟
⎝b⎠
=1
30. x = a(sec θ + tan θ) ; y = b (sec θ – tan θ)
Sol. x = a(sec θ + tan θ) ; y = b (sec θ – tan θ)
xy = ab(sec2 θ – tan2 θ) = ab (1) = ab
xy = ab.
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⎛π
⎞
cos (π − A ) cot ⎜ + A ⎟ cos ( − A )
⎝2
⎠
31. Prove that
= cos A
⎛ π
⎞
tan (π + A ) tan ⎜ 3 + A ⎟ sin ( 2π − A )
⎝ 2
⎠
SAQ’S
32. Prove that 3(sin θ − cos θ) 4 + 6(sin θ + cos θ) 2 + 4(sin 6 θ + cos6 θ) = 13
Sol. Consider (sin θ − cos θ) 2 = sin2 θ + cos2 θ – 2 sin θ cos θ= 1 – 2 sin θ cos θ
(sin θ – cos θ)4 = [(sin θ – cos θ)2]2 = [1 − 2sin θ cos θ]2 = 1 + 4sin 2 θ cos 2 θ − 4sin θ cos θ
∴ (sin θ − cos θ) 4 = 1 + 4sin 2 θ cos 2 θ − 4sin θ cos θ
(sin θ + cos θ) 2 = sin 2 θ + cos 2 θ + 2sin θ cos θ
= 1 + 2sin θ cos θ
∴ (sin θ + cos θ) 2 = 1 + 2sin θ cos θ
sin 6 θ + cos6 θ = (sin 2 θ)3 + (cos 2 θ)3
= (sin 2 θ + cos 2 θ)3 − 3sin 2 θ cos 2 θ(sin 2 θ + cos 2 θ)
= 1 − 3sin 2 θ cos 2 θ (∵ sin 2 θ + cos 2 θ = 1)
∴ sin 6 θ + cos6 θ = 1 − 3sin 2 θ cos 2 θ
∴ 3(sin θ − cos θ) 4 + 6(sin θ + cos θ) 2 + 4(sin 6 θ + cos6 θ)
= 3[1 + 4sin 2 θ cos 2 θ − 4sin θ cos θ] + 6[1 + 2sin θ cos θ] + 4[1 − 3sin 2 θ cos 2 θ]
= 3 + 12sin 2 θ cos 2 θ − 12sin θ cos θ + 6 + 12sin θ cos θ + 4 − 12sin 2 θ cos 2 θ
= 3 + 6 + 4 = 13
1 ⎤
⎡
33. Show that cos 4 α + 2 cos 2 α ⎢1 −
= 1 − sin 4 α .
2 ⎥
⎣ sec α ⎦
1 ⎤
⎡
Sol. L.H.S. : cos 4 α + 2 cos 2 α ⎢1 −
2 ⎥
⎣ sec α ⎦
= cos 4 α + 2 cos 2 α sin 2 α
(∵1 − cos 2 α = sin 2 α)
= cos 2 α[cos 2 α + 2sin 2 α]
= (1 − sin 2 α)[(1 − sin 2 α) + 2sin 2 α]
= (1 − sin 2 α)(1 + sin 2 α)
= 1 − sin 4 α = R.H.S.
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34. Prove that
(1 + sin θ − cos θ) 2 1 − cos θ
.
=
(1 + sin θ + cos θ) 2 1 + cos θ
Sol. Consider
1 + sin θ − cos θ = (1 − cos θ) + sin θ
= 2sin 2
θ
θ
θ
θ⎡ θ
θ⎤
+ 2sin cos = 2sin ⎢sin + cos ⎥
2
2
2
2⎣ 2
2⎦
Again consider
1 + sin θ + cos θ = (1 + cos θ) + sin θ
θ
θ
θ
+ 2sin cos
2
2
2
θ⎡
θ
θ⎤
= 2 cos ⎢cos + sin ⎥
2⎣
2
2⎦
= 2 cos 2
2
⎡
θ⎡ θ
θ⎤ ⎤
⎢ 2sin 2 ⎢sin 2 + cos 2 ⎥ ⎥
⎣
⎦⎥
L.H.S.: ⎢
⎢ 2 cos θ ⎡cos θ + sin θ ⎤ ⎥
⎢⎣
2 ⎢⎣
2
2 ⎥⎦ ⎥⎦
θ
θ
sin 2
2sin 2
2 =
2 = 1 − cos θ = R.H.S.
=
θ
θ 1 + cos θ
cos 2
2 cos 2
2
2
2sin θ
1 − cos θ + sin θ
35. If
.
= x , find the value of
1 + cos θ + sin θ
1 + sin θ
2sin θ
=x
Sol.
1 + cos θ + sin θ
θ
θ
2 ⋅ sin cos
2
2
⇒
=x
θ
θ
2 θ
2 cos + 2sin cos
2
2
2
θ
θ
4sin cos
2
2
⇒
=x
θ⎡
θ
θ⎤
2 cos ⎢cos + sin ⎥
2⎣
2
2⎦
θ
2sin
2
⇒
=x
...(1)
θ
θ
cos + sin
2
2
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θ
θ
θ
+ 2sin cos
2
2
2
=
θ
θ
2 θ
2 θ
sin + cos + 2sin cos
2
2
2
2
θ⎡ θ
θ⎤
θ
2sin ⎢sin + cos ⎥
2sin
2⎣ 2
2⎦
2
=
=
= x (∵ from(1))
θ
θ
θ
θ
⎡
⎤
sin + cos
⎢⎣sin 2 + cos 2 ⎥⎦
2
2
1 − cos θ + sin θ
1 + sin θ
2sin 2
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