Solution

TRIGONOMETRY RATIO AND IDENTITIES QUESTIONS BANK

JEE main - Mathematics

1.
(c) 3
Explanation: tan 20° tan 40° tan 60° tan 80°
–
= tan 20° tan (60° - 20°) ⋅√3⋅ tan (60° + 20°)
– – –
= √3 tan 60° = √3 × √3 = 3
2. (a) 3

Explanation: We have, cos2 10o - cos 10o cos 50o + cos2 50o
= [2 cos 10 − 2 cos 10 cos 50 + 2 cos 50 ]
1

2
2 ∘ ∘ ∘ 2 ∘

= 1

2
[1 + cos 20
∘
− (cos 60
∘ ∘
+ cos 40 ) + 1 + cos 100 ]
∘

[∵ 2 cos2 A = 1 + cos 2A and 2 cos A cos B = cos (A + B) + cos(A - B)]

= [2 + cos 20 + cos 100 − − cos 40 ] [∵ cos 60 = ]
1

2
∘ ∘ 1

2
∘ ∘ 1

1 3
= 2
[
2
+ (cos 20
∘
− cos 40 ) + cos 100 ]
∘ ∘

∘ ∘ ∘ ∘
20 + 40 20 − 40
= 1

2
[
3

2
− 2 sin
2
sin
2
+ cos 100 ]
∘

E_
C+D C−D
[∵ cos C − cos D = −2 sin sin ]
2 2

= 1

2
[
3

2
− 2 sin 30
∘
sin(− 10 ) + cos(90
∘ ∘
+ 10 )]
∘

3
= 1 ∘ ∘
[∵ cos(90 ∘
]
JE
[ + sin 10 − sin 10 ] + θ) = − sin θ
2 2

3 3
= 1

2
×
2
=
4

3. (a) sec A cosec A + 1

Explanation: Given expression is tan A
+
cot A
e
1−cot A 1−tan A

= sin A
×
sin A
+
cos A
×
cos A
vin

cos A sin A−cos A sin A cos A−sin A

3 3
sin A− cos A
= sin A−cos A
1
{
cos A sin A
}

2 2
sin A+sin A cos A+ cos A
= sin A cos A
1+sin A cos A
= sin A cos A
= 1 + sec A cosec A
Di

4.
(c) 2
Explanation: 2
√3−1
5. (a)
2√2

√3−1
Explanation:
2√2

6.
2 2
p −q
(c) 2 2
p +q

Explanation: sin x + cos y = p and x cos x + sin y = q

⇒ p2 + q2 = (sin2 x + cos2 x) + (cos2 y + sin2 y) + 2 sin x cos y + 2 cos x sin y
⇒ p2 + q2 = 2 + 2 sin(x + y) ...(i)
Also, p2 - q2 = (sin2 x - cos2 x) + (cos2 y - sin2 y) + 2 sin x cos y - 2 cos x sin y
⇒ p2 - q2 = - cos 2x + cos 2y + 2sin (x - y) = 2 sin (x - y) [sin (x + y) + 1]
2 2

p2 - q2 = 2 sin(x - y).(
p +q
⇒
2
) ...[From (i)]
2 2
p −q
⇒ sin(x - y) = 2 2
p +q

7.
(d) 0

1 / 11
 Explanation: One of the factor of the given expression is cos90° and cos 90° = 0
⇒ cos 1°. cos 2°. cos 3°... cos 179° = 0

8.
(b) cot α
Explanation: Add and subtract cot α ,
cot α + (tan α - cot α ) + 2 tan 2α + 4 tan 4α + 8 cot 8α
2 2
cos α− sin α
= cot α - ( sin α cos α
) + 2 tan 2α + 4 tan 4α + 8 cot 8α
= cot α - 2 cot 2α + 2 tan 2α + 4 tan 4α + 8 cot 8α
= cot α + 2 (tan 2α - cot 2α ) + 4 tan 4α + 8 cot 8α
= cot α + 2 (-2 cot 4α ) + 4 tan 4α + 8 cot 8α
= cot α + 4 (tan 4α - cot 4α ) + 8 cot 8α
= cot α + 4 (-2 cot 8α ) + 8 cot 8α
= cot α
9.
(c) 0
Explanation: a sin θ + b cos θ = a
+
b

sin θ cos θ

⇔ a sin2 θ cos θ + b cos2 θ + sin θ

= a cos θ + b sin θ
⇔ a(1 - cos2 θ) cos θ + b(1 - sin2 θ) sin θ

E_
= a cos θ + b sin θ
⇔ -(a cos3 θ + b sin3 θ) = 0
Required expression = (a cos3 θ + b sin3 θ)2
JE
=0
10.
(b) 9

16
e

Explanation: 9

16
vin

11. (a) − 1

​ 3
√

2 cos y−1
Explanation: cos x

1
=
2−cos y

1+cos x 2−cos y+2 cos y−1

⇔ =
1−cos x 2−cos y−2 cos y+1
Di

2 x
2 cos
2 1+cos y
⇔ x =
2 3−3 cos y
2 sin
2
y
2
2 cos
2 x 1 2
⇔ cot = ( y
)
2 3 2
2 sin
2

x y 1
⇒ cot( ) tan( )= ±
2 2 √3

x π π y
0 <
2
<
2
and 2
<
2
< π
y
⇒ cot
x

2
tan 2
= −
1

√3

12.
1
(d) 16

Explanation: We have, sin 10o sin 30o sin 50o sin 70o = sin (30o)[sin (10o) sin (50o) sin (70o)]
= [sin(10 ) sin(60 − 10 ) sin(60 + 10 )]
1

2
∘ ∘ ∘ ∘ ∘

= 1

2
[
1

4
sin(3 (10 ))]
∘
[∵ sin θ sin(60 ∘
− θ) sin(60
∘
+ θ) = 1

4
sin 3θ ]
= 1

8
sin 30
∘
=
1

8
×
1

2
=
1

16

13.
(c) 4
Explanation: Given expression = 4 - 4 cos2 x + 3 cos2 x
= 4 - cos2 x ≤ 4
⇒ Maximum value = 4

2 / 11
14.
(d) 6
Explanation: 6
15.
−−
(c) √19
Explanation: Given expression 3 cos θ + 5 sin(θ − π

6
)

π π
= 3 cos θ + 5 (sin θ cos 6
− sin
6
cos θ)

√3
= 3 cos θ + 5 ( 2
sin θ −
1

2
cos θ)

5√3
= 3 cos θ − 5

2
cos θ +
2
sin θ

5√3
= 1

2
cos θ +
2
sin θ
−−−−−−
∵ The maximum value of a a cos θ + b sin θ is √a 2
+ b
2

5√3
So, maximum value of 1

2
cos θ +
2
sin θ
−−−−−−−−−−−−
2
2 −−−−−− −−
5√3 −−
= √( 1

2
) + (
2
) = √
1

4
+
75

4
= √
76

4
= √19

16.
(c) 2
Explanation: 2
17.

_
2
1 √5−1
(c) ( )
2

Explanation: Let
4

cos

sin
π

20

20
= a,

= b
} ⇒ a2 + b2 = 1
EE
Then, x = a4 - b4
eJ
= (a2 - b2)(a2 + b2)
= a2 - b2
and y = a4 + b4
vin

= (a2 - b2)2 + 2a2b2

2

⇒ y = x2 + 2

4
(2 sin
π

20
cos
20
π
)

y - x2 = 1 2 π 1 2 2π
Di

⇒ sin = cos
2 10 2 5

= 1

4
(1 - cos 4π

5
)
2
√5−1
= 1

2
(
4
)

18.
(b) -1
Explanation: -1
19. (a) 0
Explanation: 0
20.
(b) − 7

Explanation: We have
5 tan2 x - 5 cos2 x = 2(2 cos2 x -1) + 9
⇒ 5 tan2 x - 5 cos2 x = 4 cos2 x - 2 + 9
⇒ 5 tan2 x = 9 cos2 x + 7
⇒ 5(sec2 x - 1) = 9 cos2 x + 7
5
Let cos 2
x = t ⇒
t
− 9t − 12 = 0

⇒ 9t2 + 12t - 5 = 0 ⇒ 9t2 + 15t - 3t - 5 = 0

⇒ (3t - 1)(3t + 5) = 0 ⇒ t = as t ≠ − 1 5

3 3

3 / 11
 cos 2x = 2 cos2x - 1 = 2 ( 1

3
) − 1 = −
1

cos2 2x
1 7
cos 4x = 2 - 1 = 2(− 3
) − 1 = −
9

21. (a) 3

Explanation: 3

22. (a) 4
Explanation: 4
23. (a) -2 sec α
−−−−− −−−−−
1−sin α 1+sin α
Explanation: √ 1+sin α
+ √
1−sin α

1−sin α+1+sin α
=
√1− sin2 α

= 2
= -2 sec α ...[∵ cos α < 0 in ( π

2
, π) ]
| cos α|

24.
(c) cot A cot B cot C = cot D
cos(A+B) sin(C−D)
Explanation: cos(A−B)
=
sin(C+D)

cos(A+B)+cos(A−B) sin(C−D)+sin(C+D)
⇔ =
cos(A+B)−cos(A−B) sin(C−D)−sin(C+D)

2 cos A cos B 2 sin C cos D

⇔ =
−2 sin A sin B −2 cos C sin D

cot A cot B cot C = cot D

_
⇔

25.
(d) (1, √2]
–

Explanation: 0 < θ < π

2
⇒ 0 < sin θ < 1 and 0 < cos θ < 1
EE
0 < sin θ < 1 ⇒ sin θ > sin2 θ ...(i)
eJ
and 0 < cos θ < 1 ⇒ cos θ > cos2 θ ...(ii)
Adding inequations (i) and (ii), we get
sin θ + cos θ > 1 ...(iii)
−−−−−−
Since, a sin θ + b cos θ ≤ √a + b
vin

2 2

–
∴ sin θ + cos θ ≤ √2 ...(iv)

(Equality occurs only when sin θ = 1

, cos θ = 1
,θ=
π

4
)
√2 √2

–
(iii) and (iv) give the set of all values of sin θ + cos θ as (1, √2]
Di

26.
1
(b) 9 2

Explanation: sin2 85° = [sin (90° - 5°)]2 = cos2 5°

⇒ sin2 5° + sin2 85° = sin2 5° + cos2 5° = 1
sin2 5° + sin2 10° + sin2 15° +... + sin2 85° + sin2 90°
= (sin2 5° + sin2 85°) + (sin2 10° + sin2 80°) + ... + (sin2 40° + sin2 50°) + sin2 90° + sin2 45°
= (1 + 1 +1 + 1 + 1 + 1+ 1+ 1)+ 1 + = 9 1

2
1

27.
(c) 1
Explanation: tan 1° tan 2° tan 3° ... tan 89°
= (tan 1° tan 89°) (tan 2° tan 88°)... (tan 44° tan 46°) tan 45°
= (tan 1° cot 1°) (tan 2° cot 2°)... (tan 44° cot 44°) ... [∵ tan (90° - θ) = cot θ]
= 1 × 1 × 1 × ... × 1 ... [∵ tan θ cot θ = 1]
=1
28.
(b) 1

2√2

Explanation: cos 3 π

8
[4 cos
3 π

8
− 3 cos
π

8
] + sin
3 π

8
[3 sin
π

8
− 4 sin
3 π

8
]

= 4 cos 6 π

8
− 4 sin
6 π

8
− 3 cos
4 π

8
+ 3 sin
4 π

4 / 11
 = 4 [(cos 2 π

8
− sin
2 π

8
)]

4 π 4 π 2 π 2 π 2 π 2 π 2 π 2 π
[(sin + cos + sin cos )] −3 [(cos − sin ) (cos + sin )]
8 8 8 8 8 8 8 8

= cos π

4
[4 (1 − sin
2 π

8
cos
2 π

8
) − 3]

= 1
[1 −
1

2
] =
1

√2 2√2

29.
∘
(b) cot(7 1

2
)

Explanation: We know that,

1+cos 2 A
cot A = sin 2 A
∘
Putting A = (7 1

2
) , we get
∘ ∘
1+cos 15
cot(7 1

2
) =
sin 15
∘

√3+1
1+
2√2
= √3−1

2√2

2√2+ √3+1 √3+1

= ×
√3−1 √3+1

– – – –
= √6 + √2 + √3 + √4
30.
(d) 16
5

Explanation: sin 36° sin72° sin 108° sin 144°

E_
= sin 36° sin 72° sin(180° - 72°) sin(180° - 36°)
= sin 36° sin 72° sin 72° sin 36° ...[∵ sin (180° - θ) = sin θ]
= sin2 36° sin2 72°
JE
2 2
√10−2√5 √10+2√5

=[ 4
] [
4
]

(10−2√5) (10+2√5)
= ⋅
e
16 4
100−4×5
= 16×16
=
16×16
80
vin

= 5

16

31.
(c) 1
Explanation: 1
Di

32.
1
(b) 2

Explanation: 1

33.
(b) − 1

Explanation: Given trigonometric equation is

2π 4π 6π
cos + cos + cos
7 7 7

2π 6π 4π
= (cos( ) + cos( )) + cos( )
7 7 7

4π 2π
= cos( ) [2 cos( ) + 1]
7 7

4π 2 π
= cos( ) [2 (1 − 2 sin ( ) + 1]
7 7

π
sin(3× )

=
4π 2 π 7 4π
= cos( ) [3 − 4 sin ( )] π
× cos( )
7 7 sin 7
7

Multiply both side by 2

3π
2 sin( )
7 4π
= π
× cos( )
2 sin 7
7

Apply 2 sin A cos B = sin(A + B) + sin(A - B)

5 / 11
 7π −π
sin( )+sin( )

Here, sin(π ) = 0
7 7
= π ,
2 sin
7
π
− sin
7 1
= π
= −
2 sin 2
7

34.
(d) 4
∘ ∘
√3 cos 10 − √3 sin 10
Explanation: sin 10
1
∘
−
cos 10
∘
=
sin 10
∘ ∘
cos 10

1 √3
∘ ∘
( cos 10 − sin 10 )
2 2

= 1 ∘ ∘
(sin 10 cos 10 )
2
∘ ∘ ∘ ∘
4(sin 30 cos 10 −cos 30 sin 10 )
= ∘
sin 20
∘ ∘
4 sin( 30 − 10 )
= sin 20
∘

∘
4 sin 20
= sin 20
∘ =4
35.
√3(1− √5)
(c) 4

Explanation: Given trigonometric equation is

sin 12o + sin 12o - sin 72o
= sin 12o - 2 cos 42o sin 30o
= sin 12o - sin (90o - 48o) = sin 12o - sin 48o

_
= -2 cos 30o sin 18o = −2 ×
√3 √5−1 √3 –
× = (1 − √5)

36.
(c) sin 1 > sin 1°
2 4 4
EE
Explanation: Since, 1 radian = 57° nearly
eJ
and sin 57° > sin 1° ⇒ sin 1 > sin 1°
37.
(b) 4
vin

Explanation: 4
38.
(b) 1

Explanation: 1
Di

39.
(d) 2
Explanation: 2
40.
4
(b) 3
1
Explanation: sin x + sin y = 2

x+y x−y
⇒ 2 sin ( 2
) cos ( 2
) =
1

2
...(i)
cos x + cos y = 1
x+y x−y
⇒ 2 sin ( 2
) cos ( 2
) = 1 ...(ii)
Dividing (i) by (ii), we get
x+y
tan ( 2
) =
1

x +y
2 tan( )

Now, tan (x + y) =
2

x +y
2
1− tan ( )
2

1
2( )

=
2 4
=
1 3
1−
4

41.
(b) 36
Explanation: Let x = π

9
or 5π

9
or 7π

6 / 11
 ⇔ 3x = π

3
or 5π

9
or 7π

⇒ cos 3x = 1

⇔ 4 cos3 x - 3 cos x = 1

Dividing by cos3 x, we get

sec3 x +6 - 8 = 0, sec2 x
which is a cubic equation in sec x whose roots
are sec , sec , sec
π

9
5π

9
7π

Let a = sec π

9
, b = sec 5π

9
, c = sec 7π

9
,
then a + b + c = -6 and ab + bc + ca = 0
Required expression = a2 + b2 + c2
= (a + b + c)2 - 2(ab + bc + ca)
= (- 6)2 - 2(0) = 36
42.
–
(d) 2(√2 + 1)
–
Explanation: 2(√2 + 1)
43.
(b) 9

Explanation: Let α = π
or 5π
or 7π

_
9 9 9

⇒ 3α =
π

3
or 5π

9
or 7π

⇔
cos 3 α =
(4 cos3 α - 3 cos α )2 -
1

2
⇒ cos2 3 α =
1

4
=0
1

4
EE
⇔ 64 cos6 α - 96 cos4 α + 36 cos2 α - 1 = 0,
eJ
which is a cubic equation in cos2 α
whose roots are cos2 π

9
, cos2 5π

9
, cos2 7π

Let a = cos2 π
, b = cos2 5π
, c = cos2 7π
,
vin

9 9 9

then a + b + c = 96

64
=
3

and ab + bc + ca = 36

64
=
9

16

Required expression = a2 + b2 + c2
Di

= (a + b + c)2 - 2(ab + bc + ca)

2

=( 3

2
) − 2(
16
9
)

= 9

4
−
9

8
=
9

44.
(b) M = 1
+
1

2
cos
π

8
2√2

Explanation: L + M = 1 − 2 sin 2 π

8
= cos
π

4
=
1
...(i)
√2

π
and L - M = − cos 8
...(ii)
From eq. (i) and (ii),
1 1 π 1 1 π 1 1 π 1 1 π
L= 2
( − cos
8
) = −
2
cos
8
and M = 2
( + cos
8
) = +
2
cos
8
√2 2√2 √2 2√2

45.
–
(c) 2√3
Explanation: 16 sin 20o sin 40° sin 80o
= 16 sin 40o sin 20o sin 80o
= 4(4 sin (60 - 20) sin(20) sin (60 + 20))
= 4 × sin (3 × 20o) [∵ sin 3θ = 4 sin(60 − θ) × sin θ × sin(60 + θ)]
= 4 × sin 60o
√3 –
= 4 × = 2√3
2

7 / 11
46.
(d) 19π

24

Explanation: tan (A - B) = 1
⇒ tan (A - B) = tan
π

⇒ A-B= π

4
...(i)
and sec (A + B) = 2

√3

⇒ sec (A + B) = sec 11π

⇒ A+B= 11π

6
...(ii)
From (i) and (ii), we get
B= 19π

24

47. (a) 3

4
3
Explanation: 4

48.
–
(c) 2√3
Explanation: tan 75° - cot 75°
= tan (90° - 15°) - cot 75°
= cot 15° - cot 75°
– –
= (2 + √3) - (2 - √3)
–
= 2√3

_
49.
(c) cosec 1o cot 1o
Explanation: cosec 1o cot 1o
EE
50. (a) 2
eJ
Explanation: 2
51. 1
Explanation:
vin

3π π
Let y = 3 [ sin 4
(
2
− α) + sin (3π + α)]
4
- 2 [sin 6
(
2
6
+ α) + sin (5π − α)] , then we have

y = 3(cos4 α + sin4 α ) - 2(cos6 α + sin6 α )

= 3(1 - 2 sin2 α cos2 α ) - 2(1 - 2 sin2 α cos2 α )
Di

= 3 - 6 sin2 α cos2 α - 2 + 6 sin2 α cos2 α = 1

52. 0.5
Explanation:
13 1
Given: cos x = 14
and cos y = 7
π
To prove: (x - y) = − 3
−−−−−−−−−−−
−−−−−−−−
−−−−−−−− − 2
13 196−169
Now, 2
sin x = √(1 − cos x) = √(1 − (
14
) ) ⇒ √(
196
)

−−−−−
27 3√3
= √( ) ⇒
196 14

−−−−−−−−−−
−−−−−− −−−−
−−−−−−−− − 2
49−1 48 4√3
2 1
sin y = √(1 − cos y) ⇒ √(1 − ( ) ) = √( ) = √( ) =
7 49 49 7

Hence, cos (x - y) = cos x.cos y + sin x.sin y

3√3 4√3 13+36
=
13

14
⋅
1

7
+
14
⋅
7
=
98
=
49

98
= 0.5
53. 1
Explanation:
NA
54. 0
Explanation:
Given: sin θ + sin θ + sin θ = 3
1 2 3

As we know that Sine function can take the maximum value of 1.

8 / 11
 If sin θ 1 + sin θ2 + sin θ3 = 3

sin θ1 = 1
π
⇒ θ1 =
2

Similarly,
π
θ2 = θ3 =
2

⇒ cos θ1 = cos θ2 = cos θ3 = 0

⇒ cos θ1 + cos θ2 + cos θ3 = 0

55. -0.8
Explanation:
We have to find the value of cos(A + B)
It is given that,
cos A = − and cos B =
24

25
3

5
−−−−−−− − −−−−−−− −
2
∴ sin A = − √1 − cos A
2
sin B = − √1 − cos B ∵ and [ In the 3rd and 4th quadrant sinθ is negative]
−−−− −−−−−− −−− −−−−−
2 2

⇒ sin A = − √1 − (−
24

25
) and sin B = −√1 − ( 3

5
)

−−−−−− −−−−−
576 9
⇒ sin A = − √1 −
625
and sin B = −√1 − 25

−−
− −−
49 16
⇒ sin A = − √
625
and sin B = −√ 25

7
⇒ sin A = −
25
and sin B = − 4

_
Now,
cos(A + B) = cos A cos B - sin A sin B
= −

= −
24

25

72
×

−
3

28
− (−
7

25
) × (−
4

5
)
EE
125 125
−72−28
=
eJ
125
−100 −4
=
125
=
5
= -0.8
56. 9
Explanation:
vin

NA
57. 2
Explanation:
NA
Di

58. 6
Explanation:
NA
59. 3
Explanation:
NA
60. 5
Explanation:
NA
61. 250
Explanation:
We have,
10 sin4 α + 15 cos4 α = 6
2
4 4 2 2
⇒ 10 sin α + 15 cos α = 6(sin α + cos α)

[Dividing both sides by cos4 α ]

2
4 2
⇒ 10 tan α + 15 = 6(tan α + 1)
2
2
⇒ (2 tan α − 3) = 0

9 / 11
 2 3
⇒ tan α =
2

∴ 27 cosec6 a + 8 sec6 a = 27 (1 + cot2 α )3 + 8 (1 + tan2 α )3

3 3

27(1 +
2

3
) + 8(1 +
3

2
) = 27 ×
125

27
+ 8 ×
125

8
= 250.

62. 253
Explanation:
NA
63. -7
Explanation:
NA
64. 2
Explanation:
NA
65. 2
Explanation:
NA
66. 4
Explanation:
4

_
67. 218
Explanation:
218
68. 23
EE
Explanation:
23
eJ
69. 80
Explanation:
∘ 1 ∘ ∘ ∘ ∘ ∘ ∘ ∘
sin 10 ( ⋅ 2 sin 20 sin 40 ) ⋅ sin 10 sin(60 − 10 ) sin (60 + 10 )
vin

2
∘ 1 ∘ ∘ 1 ∘
= sin 10 (cos 20 − cos 60 ) ⋅ sin 30
2 4
1 1 1 ∘ ∘ 1
= ⋅ ⋅ ⋅ sin 10 (cos 20 − )
2 4 2 2
1 1 ∘ ∘ ∘
= ⋅ (2 sin 10 cos 20 − sin 10 )
16 2
1 ∘ ∘ ∘
Di

= (sin 30 − sin 10 − sin 10 )

32
1 1 ∘
= − (sin 10 )
64 16

Hence α = 64
1
⇒ α
−1
= 64

Hence, 16 + α −1
= 80

70. 1
Explanation:
−−−−−
2
−
√2 sin α 1− cos β
=
1

7
and √ 2
=
10
1

√2 cos α

√2 sin β 1
⇒ =
√2 √10

∴ tan α =
1

7
and sin β = 1

√10

1
tan β =
3
1 2
2⋅
2 tan β 3 3 3
∴ tan 2β = = = =
2 1 8 4
1− tan β 1−
9 9

tan α+tan 2β
tan(α + 2β) =
1−tan α tan 2β

10 / 11
 1 3 4+21
+
7 4 28
= = = 1
1 3 25
1− ⋅
7 4 28

71. 17
Explanation:
17
72. 16
Explanation:
1
tan A ⋅ tan B =
2
1
⇒ tan A =
2 tan B

2 1
⇒ tan A=
2
4 tan B

Now, (5 − 3 cos2A)(5 − 3 cos2B)

2 2
1− tan A 1− tan B
= (5 − 3 ( 2
)) (5 − 3 (
2
))
1+ tan A 1+ tan B
2 2 2 2
5+5 tan A−3+3 tan A 5+5 tan B−3+3 tan B
=( 2
)(
2
)
1+ tan A 1+ tan B
2 2
2+8 tan A 2+8 tan B
=( 2
)(
2
)
1+ tan A 1+ tan B
2 2
1+4 tan A 1+4 tan B
=4×( 2
)(
2
)
1+ tan A 1+ tan B

1
⎛ 1+4( ) ⎞
2 2
1+4 tan B
=4×⎜
4 tan B
⎟( )
2
1+ tan B

_
1
⎝ 1+( ) ⎠
4 tan2 B

2 2
4(1+ tan B)

=4×(

= 4 × 4 = 16
1+4 tan
2
B
)(
1+4 tan

1+ tan
2
B
B
)
EE
73. 4
eJ
Explanation:
–
Given expression = √3cosec 20 ∘
− sec 20
∘ ∘ ∘
= tan 60 cosec 20
∘
− sec 20
∘ ∘ ∘ ∘
sin 60 cos 20 −cos 60 ⋅sin 20
=
vin

∘ ∘ ∘
cos 60 ⋅sin 20 ⋅cos 20
∘ ∘ ∘
sin( 60 − 20 )
= ∘
cos 60 ⋅sin 20 ⋅cos 20
∘ ∘
= 1
sin 40

∘ ∘
⋅sin 20 cos 20
2
∘ ∘
2 sin 20 cos 20
= 1 ∘ ∘
= 4
sin 20 cos 20
2

74. 4
Di

Explanation:
Given, tan9o - tan 27o - tan 63o + tan 81o
= tan 9o + cot 9o - tan 27o - cot 27o
2 2 2×4 2×4
= ∘
− ∘
= − = 4
sin 18 sin 54 √5−1 √5+1

75. 11
Explanation:
11

11 / 11