NCERT Solutions for Class 11 Maths Chapter Trigonometric Functions
Table of Contents
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)
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.
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.
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).
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:
- Which trigonometric function is best suited to calculate the height of the tower?
(A) Sine
(B) Cosine
(C) Tangent
(D) Cosecant - What is the height of the tower?
(A) 50√3 m
(B) 50 m
(C) 50/√3 m
(D) 25 m - 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 - 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:
- (C) Tangent
(Since tan θ = Height / Base, we use tan 30° = h / 50) - (C) 50/√3 m
(h = 50 × tan 30° = 50 × 1/√3 = 50/√3 m) - (A) Increases
- (B) Decreases
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:
- 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) - What is the maximum height reached by the passenger?
(A) 10 m
(B) 20 m
(C) 30 m
(D) 40 m - 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:
- (D) h(t)=20+10sin(π/20t)
(Since the wheel rotates sinusoidally with mid-height at 20m) - (B) 20 m
- (B) 20 sec
(Since one full revolution takes 40 sec, half-revolution takes 20 sec)
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