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NCERT Solutions for Class 11 Maths Chapter Trigonometric Functions

Exercise 3.1

1. Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47°30′ (iii) 240° (iv) 520°

SOLUTION

(i) Angle in radian = Angle in degree × π/180°

= 25° × π/180°

= 5π/36 radian

(ii) -47° 30′ = -47° – 30 × 1/60

= -47° – 0.5° = -47.5°

Angle in radian = Angle in degree × π/180°

= -47.5° × π/180°

= -9.5π/36 radian

(iii) Angle in radian = Angle in degree × π/180°

= 240° × π/180°

= 4π/3 radian

(iv) Angle in radian = Angle in degree × π/180°

= 520° × π/180°

= 26π/9 radian

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7).

(i) 11/16 (ii) – 4 (iii) 5π/3 (iv) 7π/6

SOLUTION

(i) Angle in degree = Angle in radian × 180°/π

= 11/16 × 180° × 7/22

= 630/16 °

= 39.375° = 39° + 0.375°

= 39° 0.375° × 60

= 39° 22.56′

= 39° 22.56′ = 39° 22′ (0.56 × 60)”

= 39° 22′ 33.6”

(ii) Angle in degree = Angle in radian × 180°/π

= (-4) × 180° × 7/22

= -5040/22 °

= -2520/11 ° = -229+(1/11)°

= -229° (1/11) × 60

= -229° + (5+5/11) × 60

= -229° 5′ (5/11) × 60

= -229° 5′ 27.27”

(iii) Angle in degree = Angle in radian × 180°/π

= 5π/3 × 180°/π

= 300°

(iv) Angle in degree = Angle in radian × 180°/π

= 7π/6 × 180°/π

= 210°

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

SOLUTION

Number of revolutions made by the wheel in one minute = 360

Number of revolutions made by the wheel in one second = 360/60 = 6

Angle (in radians) the wheel turns in one revolution = 360° = 2π

Angle (in radians) the wheel turns in 60 revolutions = 2π × 6 = 12π

Hence, the wheel turns 12π radians in one second.

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).

SOLUTION

Angle subtended by an arc at the centre of circle = θ = l/r radians

l = Length of arc

r = Radius of circle

Therefore, the required angle (in degrees) = l/r × 180/π

= 22/100 × 180 × 7/22

= 63/5 ° = 12° + (3/5)°

= 12° 3/5 × 60

= 12° 36′

Hence, the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm in degrees is 12° 36′.

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

SOLUTION

Radius of the circle = 40/2 = 20 cm

Length of chord = 20 cm

Let the centre be O and the chord be AB.

AO = BO = 20 cm (Radii of the circle)

AB = 20 cm (Given)

Therefore, AOB is an equilateral triangle.

This implies that all angles are 60°.

∠AOB = 60° = 60 × π/180 radian = π/3 radian

Let the length of minor arc AB = l

∠AOB = l/r

π/3 = l/20

20π/3 = l

Hence, the length of the required arc is 20π/3 cm.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

SOLUTION

Let there be two circles of radii r₁ and r₂ with angles θ₁ = 60° and θ₂ = 75° respectively subtended at the centre by arcs of same length l.

θ₁ = 60° × π/180° radian = π/3 radian

θ₂ = 75° × π/180° radian = 5π/12 radian

l = r₁θ₁

l = r₂θ₂

r₁θ₁ = r₂θ₂

r₁ × π/3 = r₂ × 5π/12

r₁/r₂ = 5/4

Hence, the ratio of the required radii is 5 : 4.

7. Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length

(i) 10 cm (ii) 15 cm (iii) 21 cm

SOLUTION

Radius = Length of pendulum = 75 cm

(i) Angle subtended = 10/r = 10/75

= 2/15 radian

(ii) Angle subtended = 15/r = 15/75

= 1/5 radian

(iii) Angle subtended = 21/r = 21/75

= 7/25 radian

Exercise 3.2

Find the values of other five trigonometric functions in Exercises 1 to 5.

1. cos x = -1/2, x lies in third quadrant.

SOLUTION

cos x = -1/2

sec x = 1/cos x = -2

We know that

cos² x + sin² x = 1

sin² x = 1 – cos² x

sin² x = 1 – (-1/2)²

sin² x = 1 – 1/4

sin² x = 3/4

sin x = ±√3/2

It is given that x lies in the third quadrant. Therefore, sin x will be negative.

sin x = -√3/2

cosec x = 1/sin x = -2/√3

tan x = sin x/cos x

= -√3/2 × (-2/1)

tan x = √3

cot x = 1/tan x = 1/√3

2. sin x = 3/5, x lies in second quadrant.

SOLUTION

sin x = 3/5

cosec x = 1/sin x = 5/3

We know that

cos² x + sin² x = 1

cos² x = 1 – sin² x

cos² x = 1 – (3/5)²

cos² x = 1 – 9/25

cos² x = 16/25

cos x = ±4/5

It is given that x lies in the second quadrant. Therefore, cos x will be negative.

cos x = -4/5

sec x = 1/cos x = -5/4

tan x = sin x/cos x

= 3/5 × (-5/4)

tan x = -3/4

cot x = 1/tan x = -4/3

3. cot x = 3/4, x lies in third quadrant.

SOLUTION

cot x = 3/4

tan x = 1/cot x = 4/3

We know that

1 + tan² x = sec² x

sec² x = 1 + (4/3)²

sec² x = 1 + 16/9

sec² x = 25/9

sec x = ±5/3

It is given that x lies in the third quadrant. Therefore, sec x will be negative.

sec x = -5/3

cos x = 1/sec x = -3/5

We know that

1 + cot² x = cosec² x

cosec² x = 1 + (3/4)²

cosec² x = 1 + 9/16

cosec² x = 25/16

cosec x = ±5/4

It is given that x lies in the third quadrant. Therefore, cosec x will be negative.

cosec x = -5/4

sin x = 1/cosec x = -4/5

4. sec x = 13/5, x lies in fourth quadrant.

SOLUTION

sec x = 13/5

cos x = 1/sec x = 5/13

We know that

1 + tan² x = sec² x

tan² x = sec² x – 1

tan² x = (13/5)² – 1

tan² x = 169/25 – 1

tan² x = 144/25

tan x = ±12/5

It is given that x lies in fourth quadrant. Therefore, tan x will be negative.

tan x = -12/5

cot x = 1/tan x = -5/12

We know that

sin² x + cos² x = 1

sin² x = 1 – cos² x

sin² x = 1 – (5/13)²

sin² x = 1 – 25/169

sin² x = 144/169

sin x = ±12/13

It is given that x lies in fourth quadrant. Therefore, sin x will be negative.

sin x = -12/13

cosec x = 1/sin x = -13/12

5. tan x = -5/12, x lies in second quadrant.

SOLUTION

tan x = -5/12

cot x = 1/tan x = -12/5

We know that

1 + tan² x = sec² x

sec² x = 1 + (-5/12)²

sec² x = 1 + 25/144

sec² x = 169/144

sec x = ±13/12

It is given that x lies in the second quadrant. Therefore, sec x will be negative.

sec x = -13/12

cos x = 1/sec x = -12/13

We know that

1 + cot² x = cosec² x

cosec² x = 1 + (-12/5)²

cosec² x = 1 + 144/25

cosec² x = 169/25

cosec x = ±13/5

It is given that x lies in the third quadrant. Therefore, cosec x will be positive.

cosec x = 13/5

sin x = 1/cosec x = 5/13

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°

SOLUTION

sin 765° = sin (720° + 45°)

= sin (2 × 360° + 45°)

= sin (2π + 45°)

We know that sin x repeats its values after each 2π interval. Therefore,

sin (2π + 45°) = sin 45°

= 1/√2

Hence, sin 765° = 1/√2.

7. cosec (-1410°)

SOLUTION

cosec (-1410°) = cosec (-1440° + 30°)

= cosec (-4 × 360° + 30°)

= cosec (-4π + 30°)

We know that cosec x repeats its values after each 2π interval. Therefore,

cosec (-4π + 30°) = cosec (2 × 2π + (-4π + 30°))

= cosec (30°) = 2

Hence, cosec (-1410°) = 2.

8. tan 19π/3

SOLUTION

tan 19π/3 = tan (18π/3 + π/3)

= tan (6π + π/3)

We know that tan x repeats its values after each π interval. Therefore,

tan (6π + π/3) = tan π/3 = √3

Hence, tan 19π/3 = √3.

9. sin (-11π/3)

SOLUTION

sin (-11π/3) = sin (-12π/3 + π/3)

= sin (-4π + π/3)

We know that sin x repeats its values after each 2π interval. Therefore,

sin (-4π + π/3) = sin (2 × 2π + (-4π + π/3))

= sin (π/3) = √3/2

Hence, sin (-11π/3) = √3/2.

10. cot (-15π/4)

SOLUTION

cot (-15π/4) = cot (-16π/4 + π/4)

= cot (-4π + π/4)

We know that cot x repeats its values after each π interval. Therefore,

cot (-4π + π/4) = cot (4 × π + (-4π + π/4))

= cot (π/4) = 1

Hence, cot (-15π/4) = 1.

Exercise 3.3

Prove that:

1. sin² π/6 + cos² π/3 – tan² π/4 = -1/2

SOLUTION

LHS = sin² π/6 + cos² π/3 – tan² π/4

= (1/2)² + (1/2)² – 1²

= 1/4 + 1/4 – 1

= 1/2 – 1

= -1/2 = RHS

Hence, proved.

2. 2 sin² π/6 + cosec² (7π/6) cos² π/3 = 3/2

SOLUTION

LHS = 2 sin² π/6 + cosec² (7π/6) cos² π/3

= 2 × (1/2)² + cosec² (6π/6 + π/6) × (1/2)²

= 2 × 1/4 + cosec² (π + π/6) × 1/4

= 1/2 + cosec² π/6 × 1/4

= 1/2 + 4 × 1/4

= 1/2 + 1 = 3/2 = RHS

Hence, proved.

3. cot² π/6 + cosec 5π/6 + 3tan² π/6 = 6

SOLUTION

LHS = cot² π/6 + cosec 5π/6 + 3tan² π/6

= (√3)² + cosec (6π/6 – π/6) + 3(1/√3)²

= 3 + cosec (π – π/6) + 3 × 1/3

= 4 + cosec π/6

= 4 + 2

= 6 = RHS

Hence, proved.

4. 2 sin² 3π/4 + 2 cos² π/4 + 2sec² π/3 = 10

SOLUTION

LHS = 2 sin² 3π/4 + 2 cos² π/4 + 2sec² π/3

= 2 sin² (4π/4 – π/4) + 2(1/√2)² + 2(2)²

= 2 sin² (π – π/4) + 2(1/2) + 2(4)

= 2 sin² π/4 + 1 + 8

= 2(1/√2)² + 9

= 2(1/2) + 9

= 1 + 9 = 10 = RHS

Hence, proved.

5. Find the value of:

(i) sin 75° (ii) tan 15°

SOLUTION

(i) sin 75° = sin (45° + 30°)

Using the identity sin (x + y) = sin x cos y + cos x sin y, we get:

sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

= (1/√2) × (√3/2) + (1/√2) × (1/2)

= √3/2√2 + 1/2√2

= (√3 + 1)/2√2

Hence, sin 75° = (√3 + 1)/2√2.

(ii) tan 15° = tan (45° – 30°)

Using the identity tan (x – y) = (tan x – tan y)/(1 + tan x tan y), we get:

Hence, tan 15° = (√3 – 1)/(√3 + 1).

Prove the following:

6. cos (π/4 – x) cos (π/4 – y) – sin (π/4 – x) sin (π/4 – y) = sin (x + y)

SOLUTION

LHS =

Multiply and divide by 2

= 1/2 × [2{cos (π/4 – x) cos (π/4 – y) – sin (π/4 – x) sin (π/4 – y)}]

= 1/2 × [2 cos (π/4 – x) cos (π/4 – y) – 2 sin (π/4 – x) sin (π/4 – y)]

Using the identities –

2 cos A cos B = cos (A + B) + cos (A – B)

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

we get:

1/2 × [2 cos (π/4 – x) cos (π/4 – y) – 2 sin (π/4 – x) sin (π/4 – y)]

= 1/2 × [cos (π/4 – x + π/4 – y) + cos (π/4 – x + π/4 – y)] (cos (A – B) cancels out)

= 1/2 × [2 cos (π/4 – x + π/4 – y)]

= cos (π/2 – (x + y))

We know that cos (π/2 – θ) = sin θ. Therefore,

cos (π/2 – (x + y)) = sin (x + y) = RHS

Hence, proved.

7. tan (π/4 + x)/tan (π/4 – x) = ((1 + tan x)/(1 – tan x))²

SOLUTION

LHS = tan (π/4 + x)/tan (π/4 – x)

Using the identities –

tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

tan (A – B) = (tan A – tan B)/(1 + tan A tan B)

we get:

tan (π/4 + x)/tan (π/4 – x)

= (tan π/4 + tan x)/(1 – tan π/4 tan x) × (1 + tan π/4 tan x)/(tan π/4 – tan x)

= (1 + tan x)/(1 – tan x) × (1 + tan x)/(1 – tan x)

= (1 + tan x)²/(1 – tan x)² = RHS

Hence, proved.

8. (cos (π + x) cos (-x))/(sin (π – x) cos (π/2 + x)) = cot² x

SOLUTION

LHS = (cos (π + x) cos (-x))/(sin (π – x) cos (π/2 + x))

= (-cos x) cos x/sin x (-sin x)

= cos² x/sin² x

= (cos x/sin x)²

= cot² x = RHS

Hence, proved.

9. cos (3π/2 + x) cos (2π + x) [cot (3π/2 – x) + cot (2π + x)] = 1

SOLUTION

LHS = cos (3π/2 + x) cos (2π + x) [cot (3π/2 – x) + cot (2π + x)]

Here,

cos (3π/2 + x) = cos (4π/2 – π/2 + x)

= cos (2π – (π/2 – x))

= cos (π/2 – x)

= sin x

cot (3π/2 – x) = cot (4π/2 – π/2 – x)

= cot (2π – (π/2 + x))

= -cot (π/2 + x)

= tan x

Now, LHS = sin x cos (2π + x) [tan x + cot(2π + x)]

= sin x cos x [tan x + cot x]

= sin x cos x [sin x/ cos x + cos x/sin x]

= sin x cos x [(sin² x + cos² x)/sin x cos x]

= sin² x + cos² x = 1 = RHS

Hence, proved.

10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

SOLUTION

LHS = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x

Multiply and divide by 2

sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x

= 1/2 × 2[sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x]

= 1/2 × [2 sin (n + 1)x sin (n + 2)x + 2 cos (n + 1)x cos (n + 2)x]

Using the identities –

2 cos A cos B = cos (A + B) + cos (A – B)

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

we get:

1/2 × [2 sin (n + 1)x sin (n + 2)x + 2 cos (n + 1)x cos (n + 2)x]

= 1/2 × [-{cos (n + 1 + n + 2)x – cos (n + 1 – n – 2)x} + cos (n + 1 + n + 2)x + cos (n + 1 – n – 2)x]

= 1/2 × [2 cos (n + 1 – n – 2)x] (cos (A + B) cancels out)

= 1/2 × [2 cos (-x)]

= 1/2 × 2 cos x

= cos x = RHS

Hence, proved.

11. cos (3π/4 + x) – cos (3π/4 – x) = -√2 sin x

SOLUTION

LHS = cos (3π/4 + x) – cos (3π/4 – x)

Using the identity cos A – cos B = -2 sin ((A + B)/2) sin ((A – B)/2), we get:

= -2 sin [(3π/4 + x + 3π/4 – x)/2] sin [(3π/4 + x – 3π/4 + x)/2]

= -2 sin [(3π/2)/2] sin [2x/2]

= -2 sin 3π/4 sin x

Here,

sin 3π/4 = sin (π – π/4)

= sin π/4

Now, LHS = -2 sin π/4 sin x

= -2 (1/√2) sin x

= -√2 sin x = RHS

Hence, proved.

12. sin² 6x – sin² 4x = sin 2x sin 10x

SOLUTION

LHS = sin² 6x – sin² 4x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

Using the identities –

sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2)

sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

we get:

(sin 6x + sin 4x) (sin 6x – sin 4x)

= [2 sin ((6x + 4x)/2) cos ((6x – 4x)/2)] [2 cos ((6x + 4x)/2) sin ((6x – 4x)/2)]

= [2 sin (10x/2) cos (2x/2)] [2 cos (10x/2) sin (2x/2)]

= 4 sin 5x cos x cos 5x sin x

= (2 sin 5x cos 5x)(2 sin x cos x)

Using the identity sin 2A = 2 sin A cos A, we get:

LHS = sin 10x sin 2x = RHS

Hence, proved.

13. cos² 2x – cos² 6x = sin 4x sin 8x

SOLUTION

LHS = cos² 2x – cos² 6x = (cos 2x + cos 6x) (cos 2x – cos 6x)

Using the identities-

cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2)

cos A – cos B = -2 sin ((A + B)/2) sin ((A – B)/2)

we get:

(cos 2x + cos 6x) (cos 2x – cos 6x)

= [2 cos ((2x + 6x)/2) cos ((2x – 6x)/2)] [-2 sin ((2x + 6x)/2) sin ((2x – 6x)/2)]

= [2 cos ((8x)/2) cos ((-4x)/2)] [-2 sin ((8x)/2) sin ((-4x)/2)]

= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]

= [2 cos 4x cos 2x] [2 sin 4x sin 2x]

= [2 sin 4x cos 4x] [2 sin 2x cos 2x]

Using the identity sin 2A = 2 sin A cos A, we get:

LHS = sin 2(4x) sin 2(2x)

= sin 8x sin 4x = RHS

Hence, proved.

14. sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x

SOLUTION

LHS = sin 2x + 2 sin 4x + sin 6x

= (sin 2x + sin 6x) + 2 sin 4x

Using the identity sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2), we get:

(sin 2x + sin 6x) + 2 sin 4x

= 2 sin ((2x + 6x)/2) cos ((2x – 6x)/2) + 2 sin 4x

= 2 sin ((8x)/2) cos ((-4x)/2) + 2 sin 4x

= 2 sin 4x cos (-2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

= 2 sin 4x (cos 2x + 1)

Using the identity cos 2x = 2 cos² x – 1, we get:

LHS = 2 sin 4x (2 cos² x – 1 + 1)

= 2 sin 4x (2 cos² x)

= 4 cos² x sin 4x = RHS

Hence, proved.

15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

SOLUTION

LHS = cot 4x (sin 5x + sin 3x)

= cos 4x/sin 4x × [sin 5x + sin 3x]

Using the identity sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2), we get:

cos 4x/sin 4x × [sin 5x + sin 3x]

= cos 4x/sin 4x × [2 sin ((5x + 3x)/2) cos ((5x – 3x)/2)]

= cos 4x/sin 4x × [2 sin ((8x)/2) cos ((2x)/2)]

= cos 4x/sin 4x × [2 sin 4x cos x]

= 2 cos 4x cos x

RHS = cot x (sin 5x – sin 3x)

= cos x/sin x × [sin 5x – sin 3x]

Using the identity sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2), we get:

cos x/sin x × [sin 5x – sin 3x]

= cos x/sin x × [2 cos ((5x + 3x)/2) sin ((5x – 3x)/2)]

= cos x/sin x × [2 cos ((8x)/2) sin ((2x)/2)]

= cos x/sin x × [2 cos 4x sin x]

= 2 cos 4x cos x

Since, LHS = RHS.

Hence, proved.

16. (cos 9x – cos 5x)/(sin 17x – sin 3x) = – (sin 2x)/(cos 10x)

SOLUTION

LHS = (cos 9x – cos 5x)/(sin 17x – sin 3x)

Using the identities-

cos A – cos B = -2 sin ((A + B)/2) sin ((A – B)/2)

sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

we get:

(cos 9x – cos 5x)/(sin 17x – sin 3x)

= [-2 sin ((9x + 5x)/2) sin ((9x – 5x)/2)]/[2 cos ((17x + 3x)/2) sin ((17x – 3x)/2)]

= [-2 sin ((14x)/2) sin ((4x)/2)]/[2 cos ((20x)/2) sin ((14x)/2)]

= [-2 sin 7x sin 2x]/[2 cos 10x sin 7x]

= -(sin 2x)/(cos 10x) = RHS

Hence, proved.

17. (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x

SOLUTION

LHS = (sin 5x + sin 3x)/(cos 5x + cos 3x)

Using the identities-

cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2)

sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2)

we get:

(sin 5x + sin 3x)/(cos 5x + cos 3x)

= [2 sin ((5x + 3x)/2) cos ((5x – 3x)/2)]/[2 cos ((5x + 3x)/2) cos ((5x – 3x)/2)]

= [2 sin ((8x)/2) cos ((2x)/2)]/[2 cos ((8x)/2) cos ((2x)/2)]

= [2 sin 4x cos x]/[2 cos 4x cos x]

= sin 4x/cos 4x = tan 4x = RHS

Hence, proved.

18. (sin x – sin y)/(cos x + cos y) = tan ((x – y)/2)

SOLUTION

LHS = (sin x – sin y)/(cos x + cos y)

Using the identities-

cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2)

sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

we get:

(sin x – sin y)/(cos x + cos y)

= [2 cos ((x + y)/2) sin ((x – y)/2)]/[2 cos ((x + y)/2) cos ((x – y)/2)]

= sin ((x – y)/2)/cos ((x – y)/2)

= tan ((x – y)/2) = RHS

Hence, proved.

19. (sin x + sin 3x)/(cos x + cos 3x) = tan 2x

SOLUTION

LHS = (sin x + sin 3x)/(cos x + cos 3x)

Using the identities-

cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2)

sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2)

we get:

(sin x + sin 3x)/(cos x + cos 3x)

= [2 sin ((x + 3x)/2) cos ((x – 3x)/2)]/[2 cos ((x + 3x)/2) cos ((x – 3x)/2)]

= [2 sin ((4x)/2) cos ((-2x)/2)]/[2 cos ((4x)/2) cos ((-2x)/2)]

= [2 sin (2x) cos (-x)]/[2 cos (2x) cos (-x)]

= sin 2x/cos 2x = tan 2x = RHS

Hence, proved.

20. (sin x – sin 3x)/(sin² x – cos² x) = 2 sin x

SOLUTION

LHS = (sin x – sin 3x)/(sin² x – cos² x)

Using the identity sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2), we get:

(sin x – sin 3x)/(sin² x – cos² x)

= [2 cos ((x + 3x)/2) sin ((x – 3x)/2)]/(sin² x – cos² x)

= [2 cos ((4x)/2) sin ((-2x)/2)]/(sin² x – cos² x)

= [2 cos (2x) sin (-x)]/(sin² x – cos² x)

= [-2 cos 2x sin x]/[-(cos² x – sin² x)]

Using the identity cos² x – sin² x = cos 2x, we get:

LHS = (2 cos 2x sin x)/(cos 2x)

= 2 sin x = RHS

Hence, proved.

21. (cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x

SOLUTION

LHS = (cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x)

= (cos 4x + cos 2x + cos 3x)/(sin 4x + sin 2x + sin 3x)

Using the identities-

cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2)

sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2)

we get:

[(cos 4x + cos 2x) + cos 3x]/[(sin 4x + sin 2x) + sin 3x]

= [2 cos ((4x + 2x)/2) cos ((4x – 2x)/2) + cos 3x]/[2 sin ((4x + 2x)/2) cos ((4x – 2x)/2) + sin 3x]

= [2 cos (6x)/2) cos ((2x)/2) + cos 3x]/[2 sin ((6x)/2) cos ((2x)/2) + sin 3x]

= [2 cos 3x cos x + cos 3x]/[2 sin 3x cos x + sin 3x]

= cos 3x (2 cos x + 1)/sin 3x (2 cos x + 1)

= cos 3x/sin 3x = cot 3x = RHS

Hence, proved.

22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

SOLUTION

LHS = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

Using the identity cot (A + B) = (cot A cot B – 1)/(cot A + cot B), we get:

cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cot x cot 2x – [(cot 2x cot x – 1)/(cot 2x + cot x)] × (cot 2x + cot x)

= cot x cot 2x – (cot 2x cotx – 1)

= cot x cot 2x – cot 2x cotx + 1

= 1 = RHS

Hence, proved.

23. tan 4x = (4 tan x (1 – tan² x))/(1 – 6 tan² x + tan⁴ x)

SOLUTION

LHS = tan 4x = tan 2(2x)

Using the identity tan 2A = 2 tan A/(1 – tan² A), we get:

tan 2(2x) = 2 tan 2x/(1 – tan² 2x)

Now, tan 2x = 2 tan x/(1 – tan² x). Thus,

LHS = 2 [2 tan x/(1 – tan² x)]/[1 – (2 tan x/(1 – tan² x))²]

= [4 tan x/(1 – tan² x)]/[1 – (4 tan² x/(1 – tan² x)²)]

= [4 tan x/(1 – tan² x)]/[((1 – tan² x)² – 4 tan² x)/(1 – tan² x)²]

= 4 tan x/(1 – tan² x) × (1 – tan² x)²/{(1 – tan² x)² – 4 tan² x}

= (4 tan x (1 – tan² x))/(1 + tan⁴ x – 2 tan²x – 4 tan² x)

= (4 tan x (1 – tan² x))/(1 – 6 tan² x + tan⁴ x) = RHS

Hence, proved.

24. cos 4x = 1 – 8 sin² x cos²x

SOLUTION

LHS = cos 4x = cos 2(2x)

Using the identity cos 2A = 1 – 2 sin² A, we get:

cos 2(2x) = 1 – 2 sin² 2x

Using the identity sin 2A = 2 sin A cos A, we get:

1 – 2 sin² 2x = 1 – 2 × (2 sin x cos x)²

= 1 – 2 × (4 sin² x cos² x)

= 1 – 8 sin² x cos² x = RHS

Hence, proved.

25. cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1

SOLUTION

LHS = cos 6x = cos 3(2x)

Using the identity cos 3A = 4 cos³ A – 3 cos A, we get:

cos 3(2x) = 4 cos³ 2x – 3 cos 2x

Using the identity cos 2A = 2 cos² A – 1, we get:

4 cos³ 2x – 3 cos 2x

= 4 × (2 cos² x – 1)³ – 3(2 cos² x – 1)

= 4 × [2³ cos⁶ x – 1 + 3(2 cos² x)(1)² – 3(2 cos² x)²(1)] – 6 cos² x + 3

= 4 × [8 cos⁶ x – 1 + 6 cos² x – 12 cos⁴ x] – 6 cos² x + 3

= 32 cos⁶ x – 4 + 24 cos² x – 48 cos⁴ x – 6 cos² x + 3

= 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1 = RHS

Hence, proved.

Exercise 3.4

Find the principal and general solutions of the following equations:

1. tan x = √3

SOLUTION

tan x = √3 (Given)

We know that

tan π/3 = √3

We also know that the value of tan repeats after an interval of π. Therefore,

tan (π/3) = tan (π + π/3)

= tan 4π/3

Hence, the principal solutions are x = π/3 and 4π/3.

Now, tan x = tan π/3. So,

x = nπ + π/3, where n ∈ Z

Hence, the general solution is x = nπ + π/3, where n ∈ Z.

2. sec x = 2

SOLUTION

sec x = 2 (Given)

We know that

sec π/3 = 2

We also know that the value of sec repeats after an interval of 2π. Therefore,

sec (π/3) = sec (2π – π/3)

= sec 5π/3

Hence, the principal solutions are x = π/3 and 5π/3.

Now, sec x = sec π/3. Therefore,

cos x = cos π/3

So,

x = 2nπ ± π/3, where n ∈ Z

Hence, the general solution is x = 2nπ ± π/3, where n ∈ Z.

3. cot x = -√3

SOLUTION

cot x = -√3 (Given)

We know that

cot π/6 = √3

cot (π – π/6) = -cot π/6 = -√3

and cot (π – π/6) = cot 5π/6

We also know that the value of cot repeats after an interval of π. Therefore,

cot (π – π/6) = cot (π + π – π/6)

= cot (2π – π/6)

= -cot 11π/6

Hence, the principal solutions are x = 5π/6 and 11π/6.

Now, cot x = cot 5π/6. Therefore,

tan x = tan 5π/6

So,

x = nπ + 5π/6, where n ∈ Z

Hence, the general solution is x = nπ + 5π/6, where n ∈ Z.

4. cosec x = -2

SOLUTION

cosec x = -2 (Given)

We know that

cosec π/6 = 2

cosec (π + π/6) = -cosec π/6 = -2

and cosec (π + π/6) = cosec 7π/6

Also,

cosec (2π – π/6) = -cosec π/6 = -2

and cosec (2π – π/6) = cosec 11π/6

Hence, the principal solutions are x = 7π/6 and 11π/6.

Now, cosec x = cosec 7π/6. Therefore,

sin x = sin 7π/6

So,

x = nπ + (-1)ⁿ 7π/6, where n ∈ Z

Hence, the general solution is x = nπ + (-1)ⁿ 7π/6, where n ∈ Z.

Find the general solution for each of the following equations:

5. cos 4x = cos 2x

SOLUTION

cos 4x = cos 2x

cos 4x – cos 2x = 0

Using the identity cos A – cos B = -2 sin ((A + B)/2) sin ((A – B)/2), we get:

-2 sin ((A + B)/2) sin ((A – B)/2) = 0

-2 sin ((4x + 2x)/2) sin ((4x – 2x)/2) = 0

-2 sin ((6x)/2) sin ((2x)/2) = 0

-2 sin 3x sin x = 0

sin 3x sin x = 0

Now, sin 3x = 0 or sin x = 0.

Therefore,

3x = nπ OR x = nπ, where n ∈ Z

Hence, the general solution is

x = nπ/3

OR

x = nπ

where n ∈ Z.

6. cos 3x + cos x – cos 2x = 0

SOLUTION

cos 3x + cos x – cos 2x = 0

Using cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2), we get:

2 cos ((3x + x)/2) cos ((3x – x)/2) – cos 2x = 0

2 cos ((4x)/2) cos ((2x)/2) – cos 2x = 0

2 cos 2x cos x – cos 2x = 0

cos 2x (2 cos x – 1) = 0

Now, cos 2x = 0 or 2 cos x – 1 = 0.

cos 2x = 0 or cos x = 1/2 = cos π/3

Therefore,

2x = (2n + 1)π/2 OR x = 2nπ ± π/3, where n ∈ Z

Hence, the general solution is

x = (2n + 1)π/4

OR

x = 2nπ ± π/3

where n ∈ Z.

7. sin 2x + cos x = 0

SOLUTION

sin 2x + cos x = 0

Using the identity sin 2A = 2 sin A cos A, we get:

2 sin x cos x + cos x = 0

cos x (2 sin x + 1) = 0

Now, cos x = 0 or 2 sin x + 1 = 0.

cos x = 0 or sin x = -1/2 = -sin π/6

-sin π/6 = sin (π + π/6) = sin 7π/6

Therefore,

x = (2n + 1)π/2 OR x = nπ + (-1)ⁿ 7π/6, where n ∈ Z

Hence, the general solution is

x = (2n + 1)π/2

OR

x = nπ + (-1)ⁿ 7π/6

where n ∈ Z.

8. sec² 2x = 1 – tan 2x

SOLUTION

sec² 2x = 1 – tan 2x

Using the identity sec² A = 1 + tan² A, we get:

1 + tan² 2x = 1 – tan 2x

tan² 2x + tan 2x = 0

tan 2x (tan 2x + 1) = 0

Now, tan 2x = 0 or tan 2x + 1 = 0

tan 2x = 0 or tan 2x = -1 = -tan π/4

-tan π/4 = tan (π – π/4) = tan 3π/4

Therefore,

2x = nπ + 0 OR 2x = nπ + 3π/4, where n ∈ Z

Hence, the general solution is

x = nπ/2

OR

x = nπ/2 + 3π/8

where n ∈ Z.

9. sin x + sin 3x + sin 5x = 0

SOLUTION

sin x + sin 3x + sin 5x = 0

sin x + sin 5x + sin 3x = 0

Using the identity sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2), we get:

2 sin ((x + 5x)/2) cos ((x – 5x)/2) + sin 3x = 0

2 sin ((6x)/2) cos ((-4x)/2) + sin 3x = 0

2 sin 3x cos (-2x) + sin 3x = 0

2 sin 3x cos 2x + sin 3x = 0

sin 3x (2 cos 2x + 1) = 0

Now, sin 3x = 0 or 2 cos 2x + 1 = 0

sin 3x = 0 or cos 2x = -1/2 = -cos π/3

-cos π/3 = cos (π – π/3) = cos 2π/3

Therefore,

3x = nπ OR 2x = 2nπ ± 2π/3, where n ∈ Z

Hence, the general solution is

x = nπ/3

OR

x = nπ ± π/3

where n ∈ Z.

Miscellaneous Exercise

Prove that:

1. 2 cos π/13 cos 9π/13 + cos 3π/13 + cos 5π/13 = 0

SOLUTION

LHS = 2 cos π/13 cos 9π/13 + cos 3π/13 + cos 5π/13

Using the identity cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2), we get:

2 cos π/13 cos 9π/13 + cos 3π/13 + cos 5π/13

= 2 cos π/13 cos 9π/13 + [2 cos ((3π/13 + 5π/13)/2) cos ((3π/13 – 5π/13)/2)]

= 2 cos π/13 cos 9π/13 + [2 cos ((8π/13)/2) cos ((-2π/13)/2)]

= 2 cos π/13 cos 9π/13 + [2 cos 4π/13 cos (-π/13)]

= 2 cos π/13 cos 9π/13 + 2 cos 4π/13 cos π/13

= 2 cos π/13 (cos 9π/13 + cos 4π/13)

Using the identity cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2), we get:

= 2 cos π/13 (cos 9π/13 + cos 4π/13)

= 2 cos π/13 [2 cos ((9π/13 + 4π/13)/2) cos ((9π/13 – 4π/13)/2)

= 2 cos π/13 [2 cos ((13π/13)/2) cos ((5π/13)/2)

= 2 cos π/13 2 cos π/2 cos 5π/26

We know that cos π/2 = 0

Thus, LHS = 2 cos π/13 × 0 × cos 5π/26 = 0 = RHS

Hence, proved.

2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

SOLUTION

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

= sin 3x sin x + sin² x + cos 3x cos x – cos² x

= cos 3x cos x + sin 3x sin x – (cos² x – sin² x)

Using the identities

cos A cos B + sin A sin B = cos (A – B)

cos² A – sin² A = cos 2A

we get:

cos 3x cos x + sin 3x sin x – (cos² x – sin² x)

= cos (3x – x) – (cos 2x)

= cos 2x – cos 2x = 0 = RHS

Hence, proved.

3. (cos x + cos y)² + (sin x – sin y)² = 4 cos² ((x + y)/2)

SOLUTION

LHS = (cos x + cos y)² + (sin x – sin y)²

= cos² x + cos² y + 2 cos x cos y + sin² x + sin² y – 2 sin x sin y

= [sin² x + cos² x] + [sin² y + cos² y] + 2 (cos x cos y – sin x sin y)

Using the identities

cos A cos B – sin A sin B = cos (A + B)

sin² A + cos² A = 1

we get:

[sin² x + cos² x] + [sin² y + cos² y] + 2 (cos x cos y – sin x sin y)

= 1 + 1 + 2 (cos (x + y))

= 2 + 2 (cos (x + y))

= 2 [1 + cos (x + y)]

Using the identity cos 2A = 2 cos² A – 1, we get:

2 [1 + cos (x + y)]

= 2 [1 + 2 cos² (x + y)/2 – 1]

= 2 [2 cos² (x + y)/2]

= 4 cos² ((x + y)/2) = RHS

Hence, proved.

4. (cos x – cos y)² + (sin x – sin y)² = 4 sin² (x – y)/2

SOLUTION

LHS = (cos x – cos y)² + (sin x – sin y)²

= cos² x + cos² y – 2 cos x cos y + sin² x + sin² y – 2 sin x sin y

= [sin² x + cos² x] + [sin² y + cos² y] – 2 (cos x cos y + sin x sin y)

Using the identities

cos A cos B + sin A sin B = cos (A – B)

sin² A + cos² A = 1

we get:

[sin² x + cos² x] + [sin² y + cos² y] – 2 (cos x cos y + sin x sin y)

= 1 + 1 – 2 (cos (x – y))

= 2 – 2 (cos (x – y))

= 2 [1 – cos (x – y)]

Using the identity cos 2A = 1 – 2 sin² A, we get:

2 [1 – cos (x – y)]

= 2 [1 – 1 + 2 sin² (x – y)/2]

= 2 [2 sin² (x – y)/2]

= 4 sin² ((x – y)/2) = RHS

Hence, proved.

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

SOLUTION

LHS = sin x + sin 3x + sin 5x + sin 7x

= (sin x + sin 5x) + (sin 3x + sin 7x)

Using the identity sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2), we get:

(sin x + sin 5x) + (sin 3x + sin 7x)

= [2 sin ((x + 5x)/2) cos ((x – 5x)/2)] + [2 sin ((3x + 7x)/2) cos ((3x – 7x)/2)]

= [2 sin ((6x)/2) cos ((-4x)/2)] + [2 sin ((10x)/2) cos ((-4x)/2)]

= [2 sin 3x cos (-2x)] + [2 sin 5x cos (-2x)]

= 2 sin 3x cos 2x + 2 sin 5x cos 2x

= 2 cos 2x (sin 3x + sin 5x)

Using the identity sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2), we get:

2 cos 2x (sin 3x + sin 5x)

= 2 cos 2x [2 sin ((3x + 5x)/2) cos ((3x – 5x)/2)]

= 2 cos 2x [2 sin ((8x)/2) cos ((-2x)/2)]

= 2 cos 2x [2 sin 4x cos (-x)]

= 2 cos 2x (2 sin 4x cos x)

= 4 cos x cos 2x sin 4x= RHS

Hence, proved.

6. ((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x)) = tan 6x

SOLUTION

LHS = ((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x))

Using the identities

sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2)

cos A + cos B = 2 cos ((A + B)/2) cos ((A – B)/2)

we get:

((sin 7x + sin 5x) + (sin 9x + sin 3x))

= 2 sin ((7x + 5x)/2) cos ((7x – 5x)/2) + 2 sin ((9x + 3x)/2) cos ((9x – 3x)/2)

= 2 sin ((12x)/2) cos ((2x)/2) + 2 sin ((12x)/2) cos ((6x)/2)

= 2 sin 6x cos x + 2 sin 6x cos 3x

= 2 sin 6x (cos x + cos 3x)

((cos 7x + cos 5x) + (cos 9x + cos 3x))

= 2 cos ((7x + 5x)/2) cos ((7x – 5x)/2) + 2 cos ((9x + 3x)/2) cos ((9x – 3x)/2)

= 2 cos ((12x)/2) cos ((2x)/2) + 2 cos ((12x)/2) cos ((6x)/2)

= 2 cos 6x cos x + 2 cos 6x cos 3x

= 2 cos 6x (cos x + cos 3x)

Now,

((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x))

= 2 sin 6x (cos x + cos 3x)/2 cos 6x (cos x + cos 3x)

= sin 6x/cos 6x = tan 6x = RHS

Hence, proved.

7. sin 3x + sin 2x – sin x = 4 sin x cos x/2 cos 3x/2

SOLUTION

LHS = sin 3x + sin 2x – sin x

Using the identity sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2), we get:

sin 3x + (sin 2x – sin x)

= sin 3x + [2 cos ((2x + x)/2) sin ((2x – x)/2)]

= sin 3x + [2 cos ((3x)/2) sin ((x)/2)]

= sin 3x + [2 cos ((3x)/2) sin ((x)/2)]

Using the indentity sin 2A = 2 sin A cos A, we get:

sin 3x + [2 cos ((3x)/2) sin ((x)/2)]

= 2 sin 3x/2 cos 3x/2 + [2 cos ((3x)/2) sin ((x)/2)]

= 2 cos 3x/2 (sin 3x/2 + sin x/2)

Using the identity sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2), we get:

2 cos 3x/2 (sin 3x/2 + sin x/2)

= 2 cos 3x/2 (2 sin ((3x/2 + x/2)/2) cos ((3x/2 – x/2)/2))

= 2 cos 3x/2 (2 sin ((4x/2)/2) cos ((2x/2)/2))

= 2 cos 3x/2 (2 sin x cos x/2)

= 4 sin x cos x/2 cos 3x/2 = RHS

Hence, proved.

Find sin x/2, cos x/2 and tan x/2 in each of the following:

8. tan x = -4/3, x in quadrant II

SOLUTION

It is given that x is in quadrant II. Therefore,

π/2 < x < π

Divide by 2

π/4 < x/2 < π/2

x/2 lies in quadrant I, so sin x/2, cos x/2, and tan x/2 are all positive.

We know that sec² A = 1 + tan² A, so

sec² x = 1 + tan² x

sec² x = 1 + (-4/3)²

sec² x = 1 + 16/9

sec² x = 25/9

sec x = ±5/3

cos x = ±3/5

cos x = -3/5 as x is in quadrant II.

We know that cos2A = 2 cos² A – 1, so

cos x = 2 cos² x/2 – 1

-3/5 = 2 cos² x/2 – 1

2/5 = 2 cos² x/2

1/5 = cos² x/2

cos x/2 = 1/√5

Now,

sin² x/2 + cos² x/2 = 1

sin² x/2 + (1/√5)² = 1

sin² x/2 = 1 – 1/5

sin² x/2 = 4/5

sin x/2 = 2/√5

Now,

tan x/2 = sin x/2/cos x/2

= 2/√5 × √5/1

tan x/2 = 2

9. cos x = -1/3, x in quadrant III

SOLUTION

It is given that x is in quadrant III. Therefore,

π < x < 3π/2

Divide by 2

π/2 < x/2 < 3π/4

x/2 lies in quadrant II, so cos x/2, and tan x/2 are negative whereas sin x/2 is postive.

We know that cos2A = 1 – 2 sin² A, so

cos x = 1 – 2 sin² x/2

-1/3 = 1 – 2 sin² x/2

2 sin² x/2 =2/3

sin² x/2 = 2/3

sin x/2 = √2/√3

We know that cos2A = 2 cos² A – 1, so

cos x = 2 cos² x/2 – 1

-1/3 = 2 cos² x/2 – 1

2/3 = 2 cos² x/2

1/3 = cos² x/2

cos x/2 = -1/√3

Now,

tan x/2 = sin x/2/cos x/2

= √2/√3 × -√3/1

tan x/2 = -√2

10. sin x = 1/4, x in quadrant II

SOLUTION

It is given that x is in quadrant II. Therefore,

π/2 < x < π

Divide by 2

π/4 < x/2 < π/2

x/2 lies in quadrant I, so sin x/2, cos x/2, and tan x/2 are all positive.

We know that cos² A = 1 – sin² A, so

cos² x = 1 – (1/4)²

cos²x = 1 – 1/16

cos² x = 15/16

cos x = ±√15/4

cos x = -√15/4 as x is in quadrant II.

We know that cos2A = 2 cos² A – 1, so

cos x = 2 cos² x/2 – 1

-√15/4 = 2 cos² x/2 – 1

(4 – √15)/4 = 2 cos² x/2

(4 – √15)/8 = cos² x/2

cos x/2 = √(4 – √15)/2√2

Now,

sin² x/2 + cos² x/2 = 1

sin² x/2 + (√(4 – √15)/2√2)² = 1

sin² x/2 = 1 – (4 – √15)/8

sin² x/2 = (4 + √15)/8

sin x/2 = √(4 + √15)/2√2

Now,

tan x/2 = sin x/2/cos x/2

= √(4 + √15)/2√2 × 2√2/√(4 – √15)

tan x/2 = √(4 + √15)/√(4 – √15)

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MCQs for Chapter 3 Trigonometric Functions Class 11 with Answers

These Multiple Choice Questions (MCQs) are based on NCERT Class 11 Maths Chapter – Trigonometric Functions and are important for JEE & CBSE Board Exams.

1. If sin θ and cos θ are the roots of ax² – bx + c = 0, then the relation between a, b and c will be

(a) a² + b² + 2ac = 0

(b) a² – b² + 2ac = 0

(c) a² + c² + 2ab = 0

(d) a² – b² – 2ac = 0

Correct option: (b) a² – b² + 2ac = 0

Solution:

Given that sin θ and cos θ are the roots of the equation ax² – bx + c = 0, so sin θ + cos θ = b/a

and sin θ cos θ = c/a

Consider,

(sinθ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ,

(b/a)² = 1 + 2(c/a) {using the identity sin²A + cos²A = 1}

b²/a² = 1 + (2c/a)

b² = a² + 2ac

a² – b² + 2ac = 0

2. If tan A = 1/2 and tan B = 1/3, then the value of A + B is

(a) π/6

(b) π

(c) 0

(d) π/4

Correct option: (d) π/4

Solution:

Given,

tan A = 1/2, tan B = 1/3

We know that,

tan(A + B) = (tan A + tan B)/(1 – tan A tan B)

= [(1/2) + (1/3)]/ [1 – (1/2)(1/3)]

= [(3 + 2)/6]/ [(6 – 1)/6]

= 5/5

= 1

= tan π/4

Therefore, A + B = π/4

3. The value of cos 1° cos 2° cos 3° … cos 179° is

(a) 1/√2

(b) 0

(c) 1

(d) –1

Correct option: (b) 0

Solution:

cos 1° cos 2° cos 3° … cos 179°

cos 1° cos 2° cos 3° … cos 179°

= cos 1° cos 2° cos 3° … cos 89° cos 90° cos 91° … cos 179°

= cos 1° cos 2° cos 3° … cos 89° (0) cos 91° … cos 179°

= 0 {since the value of cos 90° = 0}

4. The value of sin 50° – sin 70° + sin 10° is equal to

(a) 1

(b) 0

(c) 1/2

(d) 2

Correct option: (b) 0

Solution:

sin 50° – sin 70° + sin 10°

= sin(60° – 10°) – sin(60° + 10°) + sin 10°

Using the formulas

sin(A – B) = sin A cos B – cos A sin B

sin(A + B) = sin A cos B + cos A sin B, we get;

sin 50° – sin 70° + sin 10° = sin 60° cos 10° – cos 60° sin 10° – sin 60° + cos 10° – cos 60° sin 10° + sin 10°

= -2 cos 60° sin 10° + sin 10°

= -2 × (1/2) × sin 10° + sin 10°

= – sin 10° + sin 10°

= 0

5. The value of sin (45° + θ) – cos (45° – θ) is

(a) 2 cosθ

(b) 2 sinθ

(c) 1

(d) 0

Correct option: (d) 0

Solution:

sin (45° + θ) – cos (45° – θ)

= sin (45° + θ) – sin (90° -(45° – θ)) {since sin(90° – A) = cos A}

= sin (45° + θ) – sin (45° + θ)

= 0

Alternative method:

sin (45° + θ) – cos (45° – θ)

Using the formulas

sin(A + B) = sin A cos B + cos A sin B

cos(A – B) = cos A cos B + sin A sin B, we get;

sin (45° + θ) – cos (45°– θ) = sin 45° cos θ + cos 45° sin θ – cos 45° cos θ – sin 45° sin θ

= (1/√2) cos θ + (1/√2) sin θ – (1/√2) cos θ – (1/√2) sin θ

= 0

6. The value of tan 1° tan 2° tan 3° … tan 89° is

(a) 0

(b) 1

(c) 1/2

(d) Not defined

Correct option: (b) 1

Solution:

tan 1° tan 2° tan 3° … tan 89°

= [tan 1° tan 2° … tan 44°] tan 45°[tan (90° – 44°) tan (90° – 43°)… tan (90° – 1°)]

= [tan 1° tan 2° … tan 44°] [cot 44° cot 43°……. cot 1°] × [tan 45°]

= [(tan 1° × cot 1°) (tan 2° × cot 2°)…..(tan 44° × cot 44°)] × [tan 45°]

We know that,

tan A × cot A =1 and tan 45° = 1

Hence, the equation becomes as;

= 1 × 1 × 1 × 1 × …× 1

= 1 {As 1ⁿ = 1}

7. If α + β = π/4, then the value of (1 + tan α) (1 + tan β) is

(a) 1

(b) 2

(c) – 2

(d) Not defined

Correct option: (b) 2

Solution:

Given,

α + β = π/4

Taking “tan” on both sides,

tan(α + β) = tan π/4

We know that,

tan(A + B) = (tan A + tan B)/(1 – tan A tan B)

and tan π/4 = 1.

So, (tan α + tan β)/(1 – tan α tan β) = 1

tan α + tan β = 1 – tan α tan β

tan α + tan β + tan α tan β = 1….(i)

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

= 1 + 1 [From (i)]

= 2

8. If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of (2 cot A – 5 cos A + sin A) is equal to

(a) -53/10

(b) 23/10

(c) 37/10

(d) 7/10

Correct option: (b) 23/10

Solution:

Given that A lies in the second quadrant and 3 tan A + 4 = 0.

3 tan A = -4

tan A = -4/3

cot A = 1/tan A = -3/4

Using the identity sec²A = 1 + tan²A,

sec²A = 1 + (16/9) = 25/9

sec A = √(25/9)

sec A = -5/3 (in quadrant II secant is negative)

cos A = 1/sec A = -⅗

Using the identity sin²A + cos²A = 1,

sin A = √(1 – 9/25) = √(16/25) = 4/5 (in quadrant II sine is positive)

Now,

2 cot A – 5 cos A + sin A

= 2(-3/4) – 5(-3/5) + (4/5)

= (-3/2) + 3 + (4/5)

= (-15 + 30 + 8)/10

= 23/10

9. If for real values of x, cos θ = x + (1/x), then

(a) θ is an acute angle

(b) θ is right angle

(c) θ is an obtuse angle

(d) No value of θ is possible

Correct option: (d) No value of θ is possible

Solution:

Given,

cos θ = x + (1/x)

cos θ = (x² + 1)/x

⇒ x² + 1 = x cos θ

⇒ x² – x cos θ + 1 = 0

We know that for any real root of the equation ax² + bx + c = 0, b² – 4ac ≥ 0.

⇒ (-cos θ)² – 4 ≥ 0

⇒ cos²θ – 4 ≥ 0

⇒ cos²θ ≥ 4

⇒ cos θ ≥ ± 2

We know that -1 ≤ cos θ ≤ 1.

Hence, no value of θ is possible.

10. Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2π] is

(a) 0

(b) 1

(c) 2

(d) 3

Correct option: (c) 2

Solution:

Given,

tan x + sec x = 2 cos x

(sin x/cos x) + (1/cos x) = 2 cos x

(sin x + 1)/cos x = 2 cos x

sin x + 1 = 2 cos²x

Using the identity sin²A + cos²A = 1,

sin x + 1 = 2(1 – sin²x)

sin x + 1 = 2 – 2 sin²x

⇒ 2 sin²x + sin x − 1 = 0

⇒ (2 sin x − 1)(sin x + 1)=0

⇒ sin x = −1, 1/2

⇒ x = π/6,5π/6,3π/2 ∈ [0, 2π]

But for x = 3π/2, tan x and sec x are not defined.

Therefore, there are only two solutions for the given equation in the interval [0, 2π].

Assertion & Reason Questions – Trigonometric Functions (Class 11)

These Assertion-Reason questions are designed to test conceptual understanding of NCERT Class 11 Chapter: Trigonometric Functions.

Question 1:

Assertion (A): The sine function is an odd function.
Reason (R): For an odd function, f(−x)=−f(x) holds true for all values of xxx.

• (A) Both A and R are true, and R is the correct explanation of A.
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true, but R is false.
• (D) A is false, but R is true.

Answer: (A) Both A and R are true, and R is the correct explanation of A.
Explanation: sin⁡(−x)=−sin⁡(x), which satisfies the condition of an odd function.

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Question 2:

Assertion (A): The principal value of cos⁡⁻¹(−1) is π.
Reason (R): The principal value of the inverse trigonometric function lies within its principal range.

• (A) Both A and R are true, and R is the correct explanation of A.
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true, but R is false.
• (D) A is false, but R is true.

Answer: (A) Both A and R are true, and R is the correct explanation of A.
Explanation: The principal value of cos⁡⁻¹(x) lies in [0,π], and since cos⁡(π)=−1, the principal value is π.

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Question 3:

Assertion (A): The range of tan⁡⁻¹(x) is (−π/2,π/2).
Reason (R): The function tan⁡⁻¹(x) is one-to-one and onto in its principal branch.

• (A) Both A and R are true, and R is the correct explanation of A.
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true, but R is false.
• (D) A is false, but R is true.

Answer: (A) Both A and R are true, and R is the correct explanation of A.
Explanation: The function tan⁡−1(x) is defined for all real numbers with a range of (−π/2,π/2).

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Question 4:

Assertion (A): The period of the function sin⁡(x) is 2π.
Reason (R): The period of a function is the smallest positive value of T for which f(x+T)=f(x).

• (A) Both A and R are true, and R is the correct explanation of A.
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true, but R is false.
• (D) A is false, but R is true.

Answer: (A) Both A and R are true, and R is the correct explanation of A.
Explanation: The sine function repeats its values after every 2π, satisfying the definition of periodicity.

Case Study Based Questions – Trigonometric Functions (Class 11)

Case Study 1: Height of a Tower

A surveyor is standing at a distance of 50 meters from the base of a vertical tower. He measures the angle of elevation to the top of the tower and finds it to be 30°.

Using trigonometric functions, answer the following questions.

Questions:

1. Which trigonometric function is best suited to calculate the height of the tower?
   (A) Sine
   (B) Cosine
   (C) Tangent
   (D) Cosecant
2. What is the height of the tower?
   (A) 50√3 m
   (B) 50 m
   (C) 50/√3 m
   (D) 25 m
3. If the angle of elevation is increased, what happens to the height of the tower as observed?
   (A) Increases
   (B) Decreases
   (C) Remains the same
   (D) Cannot be determined
4. If the distance between the observer and the tower is doubled, what will happen to the angle of elevation?
   (A) Increases
   (B) Decreases
   (C) Remains the same
   (D) Becomes 90°

Answers:

1. (C) Tangent
   (Since tan θ = Height / Base, we use tan 30° = h / 50)
2. (C) 50/√3 m
   (h = 50 × tan 30° = 50 × 1/√3 = 50/√3 m)
3. (A) Increases
4. (B) Decreases

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Case Study 2: Motion of a Ferris Wheel

A Ferris wheel has a radius of 10 meters and completes one full revolution in 40 seconds. A passenger starts at the lowest point.

Questions:

1. What is the equation of height h(t)h(t)h(t) of the passenger above the ground in terms of sine function?
   (A) h(t)=10sin⁡(π/20t)
   (B) h(t)=10+10sin⁡(π/20t)
   (C) h(t)=10sin⁡(π/40t)
   (D) h(t)=20+10sin⁡(π/20t)
2. What is the maximum height reached by the passenger?
   (A) 10 m
   (B) 20 m
   (C) 30 m
   (D) 40 m
3. What is the time taken to reach the highest point from the lowest point?
   (A) 10 sec
   (B) 20 sec
   (C) 30 sec
   (D) 40 sec

Answers:

1. (D) h(t)=20+10sin⁡(π/20t)
   (Since the wheel rotates sinusoidally with mid-height at 20m)
2. (B) 20 m
3. (B) 20 sec
   (Since one full revolution takes 40 sec, half-revolution takes 20 sec)

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