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COMPOUND ANGLES
Definitions and Formulae :
1. The algebraic sum of two or more angles is called a compound a compound angle. i.e.
A + B, A – B, A + B + C, A + B – C, A – B + C, B + C – A, ……. etc. are called
compound angles.
2. If A and B are any two angles then
i) Sin(A + B) = sin A cos B + cos A sin B.
ii) Sin(A – B) = sin A cos B – cos A sin B
iii) cos(A + B) = cos A cos B – sin A sin B
iv) cos(A – B) = cos A cos B + sin A sin B.
3. If A, B, A + B, A – B are not odd multiples of π/2 then
i) tan(A + B) =
tan A + tan B
.
1 − tan A tan B
ii) tan(A – B) =
tan A − tan B
.
1 + tan A tan B
4. If A, B, A + B and A – B are not integral multiples of π, then
i) cot(A + B) =
cot A cot B − 1
cot B + cot A
ii) cot(A – B) =
cot A cot B + 1
.
cot B − cot A
5. i) sin(A + B) + sin(A – B) = 2 sin A cos B
ii) sin(A + B) – sin(A – B) = 2 cos A sin B
iii) cos(A + B) + cos(A – B) = 2 cos A cos B
iv) cos(A + B) – cos(A – B) = –2 sin A sin B
v) cos(A – B) – cos(A + B) = 2sin A sin B.
6. i) sin(A + B) sin(A – B) = sin2 A – sin2 B
= cos2 B – cos2 A
ii) cos(A + B) cos(A – B) = cos2 – sin2 B
= cos2 B – sin2 A
iii) tan(A + B) tan(A – B) =
iv) cot(A + B) cot(A – B) =
tan 2 A − tan 2 B
1 − tan 2 A tan 2 B
cot 2 Acot 2 B − 1
cot 2 B − cot 2 A
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cos θ + sin θ
cos θ − sin θ
v) tan(45o + θ) =
= cot(45o – θ) =
vi) tan(45o – θ) =
1 + tan θ
1 − tan θ
cos θ − sin θ
cos θ + sin θ
= cot(45o + θ) =
1 − tan θ
1 + tan θ
and tan(45o + θ) . tan(45o – θ) = 1.
7. i) sin(A + B + C)
= ∑(cos A cos B cos C) – sin A sin B sin C
ii) cos(A + B + C)
= cos A cos B cos C – ∑(sin A sin B sin C)
iii) tan(A + B + C) =
∑ (tanA) − π(tanA)
1 − ∑ (tanAtanB)
VSAQ’S
Simplify the following
1. cos 100° ⋅ cos 40° + sin 100° ⋅ sin 40°
Sol. L.H.S. =
= cos 100° ⋅ cos 40° + sin 100° ⋅ sin 40°
1
= cos (100° – 40°) = cos 60° =
= R.H.S.
2
⎛π
⎞
⎛π
⎞
2. tan ⎜ + θ ⎟ ⋅ tan ⎜ − θ ⎟
⎝4
⎠
⎝4
⎠
π
π
tan + tan θ tan − tan θ
4
4
⋅
Sol.
π
π
1 − tan tan θ 1 + tan tan θ
4
4
(1 + tan θ) (1 − tan θ)
=
⋅
=1
(1 − tan θ) (1 + tan θ)
3. tan 75° + cot 75°
Sol. tan 75° = 2 + 3
cot 75° = 2 − 3
∴ tan 75° + cot 75° = 2 + 3 + 2 − 3 = 4
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4. Express
Sol.
( 3 cos 25° + sin 25°)
as a sine of an angle.
2
( 3 cos 25° + sin 25°)
2
3
1
cos 25° + sin 25°
2
2
= sin 60° cos 25° + cos 60° sin 25°
=
= sin(60° + 25°) = sin 85°
5. tan θ in terms of tan α, if sin(θ + α) = cos (θ + α).
Sol. sin(θ + α) = cos (θ + α)
sin(θ + α)
=1
cos(θ + α)
tan(θ + α) = 1
tan θ + tan α
=1
1 − tan θ tan α
tan θ + tan α = 1 − tan θ tan α
tan θ + tan θ tan α = 1 − tan α
tan θ[1 + tan α] = 1 − tan α
∴ tan θ =
1 − tan α
1 + tan α
cos11° + sin11°
and θ is in the third quadrant, find θ.
cos11° − sin11°
cos11° + sin11°
Sol. tan θ =
cos11° − sin11°
⎡ sin11° ⎤
cos11° ⎢1 +
⎣ cos11° ⎥⎦
=
⎡ sin11° ⎤
cos11° ⎢1 −
⎣ cos11° ⎥⎦
1 + tan11°
=
1 − tan11°
tan 45° + tan11°
(∵ tan 45° = 1)
=
1 − tan 45° tan11°
tan θ = tan(45° + 11°)
6. If tan θ =
= tan 56°
= tan(180° + 56°)
= tan 236°
θ = 236°
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7. If 0° < A, B < 90°, such that cos A =
Sol. cos A =
5
4
and sin B = , find the value of sin (A – B).
13
5
5
4
and sin B =
13
5
P
13
12
A
Q
5
R
PQ 2 = PR 2 − QR 2
= (13) 2 − 52 = 169 − 25 = 144
PQ = 12
cos A =
5
12
,sin A =
13
13
X
5
4
Y
B
3
Z
YZ2 = XZ2 − XY 2
= 52 − 42 = 24 − 16 = 9
YZ = 3
4
3
sin B = , cos B =
5
5
sin(A − B) = sin A cos B − cos A sin B
=
12 3 5 4 36 20 36 − 20 16
× − × =
−
=
=
13 5 13 5 65 65
65
65
8. What is the value of tan 20° + tan 40° +
Sol. tan 20° + tan 40° +
3 tan 20° tan 40°?
3 tan 20° tan 40°
Consider 20° + 40° = 60°
Tan(20° + 40°) = tan 60°
tan 20° + tan 40°
= 3
1 − tan 20° tan 40°
tan 20° + tan 40° = 3(1 − tan 20° tan 40°)
tan 20° + tan 40° = 3 − 3 tan 20° tan 40°
tan 20° + tan 40° + 3 tan 20° tan 40° = 3
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9. Find the value of tan 56° – tan 11° – tan 56° tan 11°.
Sol. We have 56° – 11° = 45°
tan(56° − 11°) = tan 45°
tan 56° − tan11°
=1
1 + tan 56° tan11°
tan 56° − tan11° = 1 + tan 56° tan11°
tan 56° − tan11° − tan 56° tan11° = 1
sin(C − A)
if none of sin A, sin B, sin C is zero.
cos Csin A
sin(C − A)
sin C cos A − cos Csin A
Sol. Σ
=Σ
sin C sin A
sin Csin A
sin C cos A cos C sin A
=Σ
−
sin Csin A sin Csin A
= Σ cot A − cot C
= cot A − cot C + cot B − cot A + cot C − cot B = 0
10. Evaluate Σ
11. tan 72° = tan 18° + 2 tan 54°
Sol. 72° – 18° = 54°
Take tan on both sides
tan(72° − 18°) = tan 54°
tan 72° − tan18°
= tan 54°
1 + tan 72° tan18°
tan 72° − tan18°
= tan 54°
1 + tan(90 − 18) tan18°
tan 72° − tan18°
= tan 54°
1 + tan18° tan18°
tan 72° − tan18°
= tan 54°
1+1
tan 72° − tan18°
= tan 54°
2
tan 72° − tan18° = 2 tan 54°
tan 72° = tan18° + 2 tan 54°
12. sin 750° cos 480° + cos 120° cos 60° =
Sol. sin 750° = sin(2⋅360° + 30°) = sin 30° =
−1
2
1
2
cos 480° = cos(360° + 120°) = cos 120°
= 120° = cos(180° – 60°)
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= –cos 60° = –
cos 120° = –
1
2
1
2
1
2
∴ L.H.S. =
= sin 750° cos 480° + cos 120° cos 60°
1⎛ 1⎞ ⎛ 1⎞1
= ⎜− ⎟+⎜− ⎟
2⎝ 2⎠ ⎝ 2⎠2
cos 60° =
1 1 −2 −1
=− − =
=
= R.H.S.
4 4 4
2
⎛ 4π
⎞
⎛ 4π
⎞
13. cos A + cos ⎜
− A ⎟ + cos ⎜
+ A⎟ = 0
⎝ 3
⎠
⎝ 3
⎠
⎛ 4π
⎞
⎛ 4π
⎞
− A ⎟ + cos ⎜
+ A⎟
Sol. Consider cos ⎜
⎝ 3
⎠
⎝ 3
⎠
= cos(240° + A) + cos(240° − A)
= cos 240° cos A − sin 240° sin A
+ cos 240° cos A + sin 240° sin A
= 2 cos 240° cos A
= 2 cos(180° + 60°) cos A
= −2 cos 60° cos A
1
= −2 × cos A = − cos A
2
⎛ 2π
⎞
⎛ 2π
⎞ 3
14. cos 2 θ + cos 2 ⎜
+ θ ⎟ + cos 2 ⎜
− θ⎟ =
⎝ 3
⎠
⎝ 3
⎠ 2
⎛ 2π
⎞
⎛ 2π
⎞
Sol. cos 2 ⎜
+ θ ⎟ + cos 2 ⎜
− θ⎟
⎝ 3
⎠
⎝ 3
⎠
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= cos 2 (60° + θ) + cos 2 (60° − θ)
= [cos 60° cos θ − sin 60° sin θ]2 + [cos 60° cos θ + sin 60° sin θ]2
= 2(cos 2 60° cos 2 θ + sin 2 60° sin 2 θ)[∵ (a + b) 2 + (a − b) 2 = 2(a 2 + b 2 )]
2
⎡ 1 2
⎤
⎛ 3⎞
⎛ ⎞
2
2
⎢
= 2 ⎜ ⎟ cos θ + ⎜⎜
⎟⎟ sin θ ⎥
⎢⎝ 2 ⎠
⎥
⎝ 2 ⎠
⎣
⎦
3
⎡1
⎤
= 2 ⎢ cos 2 θ + sin 2 θ ⎥
4
⎣4
⎦
2
= [cos 2 θ + 3sin 2 θ]
4
1
3
= cos 2 θ + sin 2 θ
2
2
1
3
∴ L.H.S. = cos 2 θ + cos 2 θ + sin 2 θ
2
2
3
3
= cos 2 θ + sin 2 θ
2
2
3
= ⎡⎣cos 2 θ + sin 2 θ ⎤⎦
2
3
= (∵ cos 2 θ + sin 2 θ = 1)
2
= R.H.S.
1°
1°
− sin 2 22 .
2
2
1°
1°
Sol. Put A = sin 2 82 and B = sin 2 22 , then
2
2
1°
1°
sin 2 82 − sin 2 22
2
2
15. Evaluate sin 2 82
= sin 2 A − sin 2 B
= sin(A + B) sin(A − B)
= sin105° sin 60°
= sin(90° + 15°) sin 60°
= cos15° sin 60°
=
1+ 3 3 3 + 3
⋅
=
2 2 2
4 2
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1
1
3 +1
− sin 2 22 =
2
2
4 2
0
16. Prove that sin 2 52
0
⎛π A⎞
⎛π A⎞
17. sin 2 ⎜ + ⎟ − sin 2 ⎜ − ⎟
⎝8 2 ⎠
⎝8 2 ⎠
⎛π A⎞
⎛π A⎞
Sol. sin 2 ⎜ + ⎟ − sin 2 ⎜ − ⎟
⎝8 2 ⎠
⎝8 2 ⎠
[∵ sin 2 A − sin 2 B = sin(A + B) sin(A − B)]
⎛π A π A⎞ ⎛π A π A⎞
= sin ⎜ + + − ⎟ sin ⎜ + − + ⎟
⎝8 2 8 2 ⎠ ⎝8 2 8 2 ⎠
⎛ 2π ⎞ ⎛ 2A ⎞
= sin ⎜ ⎟ sin ⎜
⎟
⎝ 8 ⎠ ⎝ 2 ⎠
1
π
= sin sin A = sin 45° sin A =
sin A
4
2
1
1
18. cos 2 52 ° − sin 2 22 °
2
2
1
1
Sol. cos 2 52 ° − sin 2 22 °
2
2
[∵ cos 2 A − sin 2 B = cos(A + B) cos(A − B)]
1 ⎞
1 ⎞
⎛ 1
⎛ 1
= cos ⎜ 52 ° + 22 ° ⎟ cos ⎜ 52 ° − 22 ° ⎟
2 ⎠
2 ⎠
⎝ 2
⎝ 2
= cos 75° cos 30°
=
3
(cos 75°)
2
=
3 ⎛ 3 −1 ⎞ 3 − 3
⎜
⎟=
2 ⎜⎝ 2 2 ⎟⎠ 4 2
0
0
0
0
0
0
⎛
⎛
1
1
1
1 ⎞
1
1 ⎞
2
19. cos 112 − sin 52 = cos ⎜112 + 52 ⎟ cos ⎜112 − 52 ⎟
2
2
2
2 ⎠
2
2 ⎠
⎝
⎝
2
(∵ cos
2
A − sin 2 B = cos ( A + B ) cos ( A − B ) )
1
1
= cos1650. cos 600 = cos (1800 − 150 ) = − cos150
2
2
1 ⎧⎪ 3 + 1 ⎫⎪
=− ⎨
⎬
2 ⎩⎪ 2 2 ⎭⎪
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20. Prove that tan 3 A tan 2 A tan A = tan 3 A − tan 2 A −tan A
Solution:
We know that tan 3 A = tan ( 2 A + A )
tan 2 A + tan A
1 − tan 2 A tan A
tan 3 A − tan 2 A tan A tan 3 = tan 2 A + tan A
tan 3 A − tan 2 − tan A = tan A tan 2 A tan 3 A
tan 3 A =
21. Find the expansion of sin(A + B + C).
Sol. sin(A + B + C)
= sin[(A + B) − C]
= sin(A + B) cos C − cos(A − B) sin C
= (sin A cos B + cos A sin B) cos C − [cos(A − B) cos B − sin A sin B]sin c
= sin A cos Bcos C + cos A sin Bcos C − cos A cos Bsin C + sin A sin Bsin C
22. Find the expansion of cos (A – B – C).
Sol. cos (A – B – C) = cos[(A – B) – C]
= cos(A − B) cos C + sin(A − B) sin C
= (cos A cos B + sin A sin B) cos C + (sin A cos B − cos A sin B]sin C
= cos A cos Bcos C + sin A sin Bcos C + sin A cos Bsin C − cos A sin Bsin C
23. For what values of x in the first quadrant
Sol.
2 tan x
is positive?
1 − tan 2 x
2 tan x
> 0 ⇒ tan 2x > 0
1 − tan 2 x
π
⇒ 0 < 2x < (since x in the first quadrant)
2
π
⇒0<x<
4
Therefore,
π
2 tan x
is positive for 0 < x <
2
4
1 − tan x
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SAQ’S
24. Prove that sin 4
Solution:
sin 4
π
8
+ sin 4
π
8
+ sin 4
3π
5π
7π 3
+ sin 4
+ sin 4
=
8
8
8
2
3π
5π
7π
+ sin 4
+ sin 4
8
8
8
2
4
π ⎞⎫
⎛ π ⎞ ⎧ 2 π π ⎫ ⎧ 2 ⎛ π π ⎞⎫ ⎧ ⎛
− ⎬ + ⎨sin ⎜ + ⎟ ⎬ + ⎨sin ⎜ π − ⎟ ⎬
⎜ sin ⎟ + ⎨sin
8⎠ ⎩
2 8⎭ ⎩
8 ⎠⎭
⎝
⎝ 2 8 ⎠⎭ ⎩ ⎝
π
π
π
π
sin 4 + cos 4 + cos 4 + sin 4
8
8
8
8
2
π
π ⎪⎫
⎪⎧⎛ 2 π
⎧ 4π
4 π ⎫
2 π ⎞
+ cos ⎬ = 2 ⎨⎜ sin
+ cos ⎟ − 2sin 2 cos 2 ⎬
2 ⎨sin
8
8⎭
8
8⎠
8
8 ⎪⎭
⎩
⎪⎩⎝
π
π
= 2 − 4sin 2 cos 2
8
8
2
π
π⎞
π
1 3
⎛
= 2 − ⎜ 2sin cos ⎟ = 2 − sin 2 = 2 − =
8
8⎠
4
2 2
⎝
4
4
5π
7π 3
+ cos 4
=
8
8
8
8
2
1
26. Prove that (i) sin A sin ( 60 − A ) sin 600 + A = sin 3 A
4
1
(ii) cos A cos ( 60 + A ) cos 600 − A = cos 3 A
4
0
0
(iii) tan A tan ( 60 + A ) tan ( 60 − A ) = tan 3 A
25. Prove that cos 4
π
+ cos 4 3
π
+ cos 4
(
)
(
)
3
16
π
2π
3π
4π 1
cos
cos
=
(v) cos cos
9
9
9
9 16
(iv) sin 200 sin 400 sin 600sin 800 =
27. Prove that tan θ + 2 tan 2θ + 4 tan 4θ + 8cot 8θ = cot θ
π
−3
7
π
and sin β =
, where < α < π and 0 < β < , then find the values of
5
25
2
2
tan(α + β) and sin(α + β).
−3
π
, where < α < π
Sol. cos α =
5
2
α in II quadrant
7
π
sin β =
, where 0 < β <
25
2
β in I Quadrant
28. If cos α =
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A
5
4
α
B
3
C
AB2 = AC2 − BC2
= 52 − 32 = 25 − 9 = 16
∴ AB = 4
3
4
4
cos α = − ,sin α = , tan α = −
5
5
3
X
25
7
β
Y
24
Z
YZ2 = XZ2 − XY 2
= 252 − 7 2
= 625 − 49 = 576
YZ = 24
7
24
7
, cos β = , tan β =
25
25
24
tan α + tan β
∴ tan(α + β) =
1 − tan α tan β
sin β =
−4 7
−32 + 7
+
4
= 3 24 =
4 7
7
1+ ×
1+
3 24
18
−25
3
−25 18
= 24 =
× =−
18 + 7
24 25
4
18
sin(α + β) = sin α cos β + cos α sin β
4 24 −3 7
= × + ×
5 25 5 25
96 21 75 3
=
−
=
=
125 125 125 5
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29. If 0 < A < B <
π
24
4
and sin(A + B) =
and cos(A − B) = , then find the value of
4
25
5
tan 2A.
Sol. sin(A + B) =
24
25
25
24
A+B
7
24
7
4
cos(A − B) =
5
tan(A + B) =
5
3
A–B
4
3
4
Now 2A = (A + B) + (A – B)
tan 2A = tan [ (A + B) + (A − B) ]
tan(A − B) =
=
tan(A + B) + tan(A − B)
1 − tan(A + B) tan(A − B)
24 3
+
96 + 21 −117
= 7 4 =
=
24 3 28 − 72
44
1− ×
7 4
30. If A + B, A are acute angles such that sin ( A + B ) =
24
3
and tan A = find the value of
25
4
cos B
Solution:
sin ( A + B ) =
24
25
∴cos ( A + B ) =
7
25
3
3
4
∴ sin A =
&
cos A =
4
5
5
cos B = cos ( A + B − A ) = cos ( A + B ) cos A + sin ( A + B ) sin A
tan A =
7 4 24 3 100 4
× + × =
=
25 5 25 5 125 5
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31. If tan α – tan β = m and cot α – cot β = n, then prove that cot(α − β) =
1 1
− .
m n
Sol. We have tan α – tan β = m
1
1
−
=m
cot α cot β
cot β − cot α
=m
cot α cot β
cot α cot β
1
=
cot β − cot α m
cot α − cot β = n
∴
...(1)
−(cot β − cot α) = n
cot β − cot α = −n
1
1
=−
cot β − cot α
n
L.H.S. = cot(α − β) =
...(2)
cot α cot β
cot β − cot α
=
cot α cot β
1
+
cot β − cot α cot β − cot α
=
1 1
− (∵ from(1) & (2)) = R.H.S.
m n
32. If tan(α – β) =
7
4
and tan α = , where α and β are in the firs quadrant prove that
24
3
α + β = π/2.
7
4
Sol. tan(α – β) =
and tan α =
24
3
7
25
2β
34
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tan(α − β) =
tan α − tan β
1 + tan α tan β
4
− tan β
7
⇒ 3
=
4
1 + tan β 24
3
4 − 3 tan β
7
3
⇒
=
3 + 4 tan β 24
3
4 − 3 tan β 7
⇒
=
3 + 4 tan β 24
⇒ 24[4 − 3 tan β] = 7(3 + 4 tan β)
⇒ 96 − 72 tan β = 21 + 28 tan β
⇒ 96 − 21 = 28 tan β + 72 tan β
⇒ 100 tan β = 75
75 3
=
100 4
tan α + tan β
∴ tan(α + β) =
1 − tan α tan β
∴ tan β =
4 3
+
3
4 =α
=
4 3
1− ⋅
3 4
π
tan(α + β) = tan
2
π
∴α + β =
2
33. If
sin(α + β) a + b
, then prove that a tan β = b tan α .
=
sin(α − β) a − b
sin(α + β) a + b
=
sin(α − β) a − b
By using componendo and dividendo, we get
sin(α + β) + sin(α − β) a + b + a − b 2a a
=
=
=
sin(α + β) − sin(α − β) a + b − a + b 2b b
Sol. Given that
⇒
sin(α + β) + sin(α − β) a
=
sin(α + β) − sin(α − β) b
⇒
sin α cos β + cos α sin β + sin α cos β − cos α sin β a
=
sin α cos β + cos α sin β − sin α cos β + cos α sin β b
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2sin α cos β a
=
2 cos α sin β b
a
⇒ tan α cot β =
b
⇒ b tan α = a tan β
hence, a tan β = b tan α
⇒
3π
, then show that
4
(1 – tan A) (1 + tan B) = 2.
3π
Sol. A − B =
4
A − B = 135°
34. If A − B =
tan(A − B) = tan135°
= tan(90° + 45)° = − cot 45° = −1
∴
tan A − tan B
= −1
1 + tan A tan B
tan A − tan B = −(1 + tan A tan B)
tan A − tan B = −1 − tan A tan B
tan A − tan B + tan A tan B = −1
tan B − tan A − tan A tan B = 1 ...(1)
L.H.S. = (1 − tan A)(1 + tan B)
= 1 + (tan B − tan A − tan A tan B)
= 1 + 1 (∵ from(1))
= 2 = R.H.S.
π
and if none of A, B, C is an odd multiple of π/2, then prove that
2
cot A + cot B + cot C = cotA cotB cotC.
π
Sol. A + B + C =
2
π
A+B = −C
2
⎛π
⎞
cot(A + B) = cot ⎜ − C ⎟
⎝2
⎠
cot A cot B − 1
= tan C
cot B + cot A
cot A cot B − 1
1
=
cot B + cot A cot C
cot C[cot A cot B − 1] = cot B + cot A
35. If A + B + C =
cot A cot Bcot C − cot C cot A + cot B
cot A cot Bcot C = cot A + cot B + cot C
∴ cot A + cot B + cot C = cot A cot Bcot C
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π
and if none of A, B, C is an odd multiple of π/2, then prove that,
2
tan A tan B + tan B tan C + tan C tan A = 1.
π
Sol. A + B + C =
2
π
A+B = −C
2
⎛π
⎞
tan(A + B) = tan ⎜ − C ⎟
⎝2
⎠
36. If A + B + C =
tan A + tan B
= cot C
1 − tan A tan B
tan A + tan B
1
⇒
=
1 − tan A tan B tan C
⇒ tan C[tan A + tan B] = 1 − tan A tan B
⇒
⇒ tan C tan A + tan C tan B = 1 − tan A tan B
⇒ tan A tan B + tan B tan C + tan C tan A = 1
cos(B + C)
= 2.
cos Bcos C
cos(B + C)
Sol. L.H.S. = Σ
cos Bcos C
cos Bcos C − sin Bsin C
=Σ
cos Bcos C
cos Bcos C sin Bsin C
=Σ
−
cos Bcos C cos B cos C
= Σ(1 − tan B tan C)
37. Σ
= 1 − tan B tan C + 1 − tan C tan A +
1 − tan A tan B
= 3 − (tan A tan B + tan B tan C + tan C tan A)
= 3 − 1 (∵ from(b))
= 2 = R.H.S.
38. Prove that sin2 α + cos2 (α + β) + 2 sin α sin β cos(α + β) is independent of α.
Sol. Given expression,
sin2α + cos2 (α+β) + 2 sinα sinβ cos(α+β)
= sin 2 α + 1 − sin 2 (α + β) + 2sin α sin β cos(α + β)
= 1 + [sin 2 α − sin 2 (α + β)] + 2sin α sin β cos(α + β)
= 1 + sin(α + α + β) sin(α − α − β) + 2sin α sin β cos(α + β)
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= 1 + sin(2α + β) sin(−β) + 2sin α sin β cos(α + β)
= 1 − sin(2α + β) sin β + [2sin α cos(α + β)]sin β
= 1 − sin(2α + β) sin α + [sin(α + α + β) + sin(α − α − β)]sin β
= 1 − sin(2α + β) sin α + [sin(2α + β) − sin β]sin β
= 1 − sin(2α + β) sin α + sin(2α + β) sin β − sin 2 β
= 1 − sin 2 β = cos 2 β
Thus the given expression is independent of α.
π
2π
3π
7π
⋅ cot
⋅ cot ...cot
= 1.
16
16
16
16
π
2π
3π
7π
Sol. cot ⋅ cot
⋅ cot ...cot
16
16
16
16
7π ⎞ ⎛
2π
6π ⎞ ⎛
3π
5π ⎞
4π
π
⎛
= ⎜ cot ⋅ cot ⎟ ⎜ cot
⋅ cot ⎟ ⎜ cot ⋅ cot ⎟ ⋅ cot
16 ⎠ ⎝
16
16 ⎠ ⎝
16
16 ⎠
16
⎝ 16
39. Prove that cot
⎡
π
π
⎛ π π ⎞ ⎤ ⎡ 2π
⎛ π 2π ⎞ ⎤ ⎡ 3π
⎛ π 3π ⎞ ⎤
= ⎢ cot ⋅ cot ⎜ − ⎟ ⎥ ⎢cot
⋅ cot ⎜ − ⎟ ⎥ ⎢cot ⋅ cot ⎜ − ⎟ ⎥ ⋅ cot
4
⎝ 2 16 ⎠ ⎦ ⎣ 16
⎝ 2 16 ⎠ ⎦ ⎣ 16
⎝ 2 16 ⎠ ⎦
⎣ 16
π
π ⎞⎛
2π
2π ⎞
⎛
= ⎜ cot ⋅ tan ⎟ ⎜ cot
⋅ tan ⎟
16 ⎠ ⎝
16
16 ⎠
⎝ 16
= 1× 1× 1× 1 = 1
3π
3π ⎞
⎛
⎜ cot ⋅ tan ⎟ ⋅1
16
16 ⎠
⎝
40. Prove that tan 70° – tan 20° = 2 tan 50°.
Sol. tan 50° = tan(70° – 20°)
tan 70° − tan 20°
=
1 + tan 70° tan 20°
⇒ tan 70° − tan 20°
= tan 50°(1 + tan 70°⋅ tan 20°)
= tan 50°(1 + tan 70°⋅ tan(90° − 70°)]
= tan 50°[1 + tan 70° ⋅ cot 70°]
= tan 50°[1 + 1]
= 2 tan 50°
∴ tan 70° − tan 20° = 2 tan 50°
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41. If A + B = 45°, then prove that
i) (1 + tan A) (1 + tan B) = 2
ii) (cot A – 1)(cot B – 1) = 2.
Sol. i) A + B = 45°
⇒ tan(A + B) = tan 45° = 1
tan A + tan B
=1
1 − tan A tan B
⇒ tan A + tan B = 1 − tan A tan B
⇒
⇒ tan A + tan B + tan A tan B = 1...(1)
Now, (1 + tan A)(1 + tan B) = 1 + tan A + tan B + tan A tan B = 2
(from(1))
ii) A + B = 45° ⇒ cot(A + B) = cot 45° = 1
cot A cot B − 1
⇒
=1
cot B + cot A
⇒ cot A cot B − 1 = cot A + cot B
⇒ cot A cot B − cot A − cot B = 1...(2)
Now, (cot A − 1)(cot B − 1) = cot A cot B − cot A − cot B + 1 = 2
(from(2))
42. If A, B, C are the angles of a triangle and if none of them is equal to π/2, then prove
that i) tan A + tan B + tan C = tan A tan B tan C
ii) cotA cotB + cotB cotC + cot C cot A =1
Sol. i) Given A + B + C = π
⇒ A+B = π−C
⇒ tan(A + B) = tan(π − C)
tan A + tan B
= − tan C
1 − tan A tan B
⇒ tan A + tan B = − tan C(1 − tan A tan B)
⇒
⇒ tan A + tan B = − tan C + tan A tan B tan C
⇒ tan A + tan B + tan C = tan A tan B tan C
1
etc., in (i) above, we get
ii) Replacing tan A by
cot A
1
1
1
1
+
+
=
cot A cot B cot C cot A cot Bcot C
⇒ cot A cot B + cot Bcot C + cot C cot A = 1
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LAQ’S
43. Let ABC be a triangle such that
cot A + cot B + cot C = 3 . Then prove that ABC is an equilateral triangle.
Sol. Given that A + B + C = 180°
We get Σ cot A cot B = 1
Now, Σ(cot A – cot B)2 = Σcot2 A + cot2 B – 2 cot A cot B
= 2 cot 2 A + 2 cot 2 B + 2 cot 2 C − 2 cot A cot B −2 cot Bcot C − 2 cot C cot A
(on expanding)
= 2{(cot A + cot B + cot C) 2 − 2(cot A cot B) − 2 cot Bcot C − 2 cot C cot A}
−2(cot A cot B + cot Bcot C + cot C cot A)
= 2(cot A + cot B + cot C) 2 − 6(cot A cot B + cot Bcot C + cot C cot A)
= 2⋅3 − 6 = 0
⇒ cot A = cot B = cot C
⇒ cot A = cot B = cot C =
3
1
=
3
3
(since cot A + cot B + cot C = 3)
⇒ A + B + C = 60°
(Since each angle lies in the interval [0,180°]
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