MP Board Class 11th Maths Important Questions Chapter 3 Trigonometric Functions

MP Board Class 11th Maths Important Questions Chapter 3 Trigonometric Functions

Trigonometric Functions Important Questions

Trigonometric Functions Objective Type Questions

(A) Choose the correct option :
Question 1.
The value of 1 + cosθ is :
(a) 2sin2 θ
(b) $$\frac { { sin }^{ 2 }\theta }{ 2 }$$
(c) 2cos2 θ
(d) cos2 θ
(c) 2cos2 θ

Question 2.
The value of $$\frac { cos 11° + sin 11° }{ cos 11° – sin 11° }$$ is :
(a) cot 56°
(b) cot 34°
(c) tan 34°
(d) tan 56°
(d) tan 56°

Question 3.
The value of sin 18° is :

Question 4.
The value of cos 1° cos 2° cos 3° ………… cos 179° is :
(a) 0
(b) 1
(c) – 1
(d) None of these
(a) 0

Question 5.
The value of $$\frac { 3 π }{ 2 }$$ radian in degree.
(a) 120°
(b) 170°
(c) 220°
(d) 270°
(d) 270°

Question 6.
The value of cos2 (60 + α) + cos2 (60 – α) + cos2 α = ……………
(a) 3
(b) $$\frac { 3 π }{ 2 }$$
(c) $$\frac { 1 }{ 2 }$$
(d) $$\frac { 3 }{ 4 }$$
(b) $$\frac { 3 π }{ 2 }$$

Question 7.
Amplitude of the function f(x) = 2 sin x is :
(a) π
(b) 2π
(c) 1
(d) 2
(d) 2

Question 8.
Period of the function f(x) = tan x is :
(a) π
(b) 2
(c) $$\frac { π }{ 2 }$$
(d) – π
(a) π

Question 9.
Solution of the equation 4 sin2θ = 1 is :
(a) nπ ± $$\frac { π }{ 3 }$$, n ∈ I
(b) 2nπ ± $$\frac { π }{ 3 }$$, n ∈ I
(c) nπ ± $$\frac { π }{ 6 }$$, n ∈ I
(d) 2nπ ± $$\frac { π }{ 6 }$$, n ∈ I
(c) nπ ± $$\frac { π }{ 6 }$$, n ∈ I

Question 10.
Maximum value of 3cosθ + 4sinθ is :
(a) 3
(b) 4
(c) 5
(d) 7
(c) 5

Question 11.
If tanθ = – $$\frac { 4 }{ 3 }$$ , then the value of sinθ is :
(a) – $$\frac { 4 }{ 5 }$$ but not $$\frac { 4 }{ 5 }$$
(b) $$\frac { 4 }{ 5 }$$ but not $$\frac – { 4 }{ 5 }$$
(c) – $$\frac { 4 }{ 5 }$$ or $$\frac { 4 }{ 5 }$$
(d) None of these
(c) – $$\frac { 4 }{ 5 }$$ or $$\frac { 4 }{ 5 }$$

Question 12.
The value of tan15° + cot15° is :
(a) 1
(b) 3
(c) 2
(d) 4
(d) 4

Question 13.
The value of sin50° + sin70° + sin10° is :
(a) 0
(b) 1
(c) – 1
(d) None of these
(a) 0

Question 14.
One value of θ which satisfy the equation cosθ +\/3sinθ = 2 is :
(a) $$\frac { π }{ 2 }$$
(b) $$\frac { π }{ 3 }$$
(c) $$\frac { 2π }{ 3 }$$
(d) $$\frac { π }{ 4 }$$
(b) $$\frac { π }{ 3 }$$

Question 15.
The general value of θ satisfying the equation cosθ = , tanθ = 1is :
(a) 2nπ + $$\frac { 5π }{ 4 }$$
(b) 2nπ – $$\frac { 5π }{ 4 }$$
(c) 2nπ + $$\frac { π }{ 4 }$$
(d) 2nπ – $$\frac { π }{ 4 }$$
(a) 2nπ + $$\frac { 5π }{ 4 }$$

(B) Match the following :

1. (d)
2. (c)
3. (a)
4. (b)
5. (g)
6. (j)
7. (e)
8. (i)
9. (h)
10. (f)

(C) Fill in the blanks :

1. The value of cos 18° = ……………….
2. The value of sin 75° = ……………….
3. If tan A = $$\frac { 5 }{ 6 }$$ and tan B = $$\frac { 1 }{ 3}$$, then the value of A + B = ……………….
4. 2π radian is equal to ………………. right angle.
5. If sinθ + cosθ = 1, then the value of sinθ.cosθ = ……………….
6. The value of $$\frac { 3tanA-{ tan }^{ 3 }A }{ 1-3{ tan }^{ 2 }A }$$ = ……………….
7. Solution of the equation cos2θ = cos2θ is ……………….
8. Amplitude of 3cosx is ……………….
9. The value of cot 22$$\frac { 1 }{ 2 }$$° is ……………….
10. If tan θ tan 2θ = 1, then the value of θ is ……………….
11. The value of sin(A + B).sin(A – B) = ……………….
12. The length of the arc of a circle of radian 6 cm substending an angle of 30° at the centre of the circle is ……………….

1. $$\frac { \sqrt { 10+2\sqrt { 5 } } }{ 4 }$$
2. $$\frac { \sqrt { 3 } + 1 }{ 2\sqrt { 2 } }$$
3. 45°
4. 4
5. 0
6. tan 3A
7. 3
8. $$\sqrt {2}$$ + 1
9. (n + $$\frac { 1 }{ 3}$$)$$\frac { π }{ 3 }$$
10. sin2A – sin2B
11. πcm

(D) Write true / false :

1. The value of tan 105° is $$\frac { \sqrt { 3 } + 1 }{ 2\sqrt { 2 } }$$
2. The value of sin 3A is 4 sin3 A + 3sinA
3. The value of cos2 A – sin2B is cos(A + B) cos (A – B)
4. The value of cos2 48°- sin2 12 is $$\frac { \sqrt { 5 } + 1 }{ 4 }$$.
5. The value of cos2 ($$\frac { π }{ 6 }$$ + θ) + sin2 ($$\frac { π }{ 6 }$$ – θ) is 0.
6. If x is a real, then the equation sinθ = x + $$\frac { 1 }{ x }$$ has unique solution.
7. Solution of tan2 θ + cot2 θ = 2 is 2nπ + $$\frac { π }{ 6 }$$
8. If f(x) = sin2 x, then f(- x) = sin2 x
9. The value of sin$$\frac { 5π }{ 12 }$$.cos$$\frac { π }{ 12 }$$ is $$\frac { \sqrt { 3 } – 2 }{ 4 }$$.
10. If A + B = $$\frac { π }{ 3 }$$ and cosA + cosB = 1, then cos(A – B) = – $$\frac { 1 }{ 3 }$$.

1. False
2. False
3. True
4. True
5. False
6. False
7. False
8. True
9. False
10. True

(E) Write answer in one word / sentence :

1. Find the solution of equation cos2 θ – sin2 θ – $$\frac { 1 }{ 4 }$$ = 0
2. The value of tan 1° tan 2° tan 3° ……………. tan 18° is :
3. The value of sin(A + B) + sin(A – B) is :
4. If (1 + tan x) (1 + tan y) = 2, then find the value of (x + y):
5. The value of sin( $$\frac { π }{ 3 }$$ + x) – sin($$\frac { π }{ 4 }$$ – x)

1. θ = nπ + (-1)n$$\frac { π }{ 6 }$$
2. 1
3. 2sinA.cosB
4. $$\frac { π }{ 4 }$$
5. $$\sqrt { 2 }$$ sinx

Trigonometric Functions Short Answer Type Questions

Question 1.
A wheel makes 360 revolutions in 1 min. then how many radians measure of an angle does it turn in 1 second? (NCERT)
Solution:
In 60 seconds number of revolutions of wheel = 360
∴ In 1 second number of revolutions of wheel = $$\frac { 360 }{ 60 }$$
∴ In one revolution angle made = 360° = 2π
∴ In 6 revolutions angle made = 2π x 6 = 12π radian

Question 2.
Find the degree measure of an angle substended at the centre of a circle of radius 100 cm by an arc of length 22 cm.
Solution:
We know that, θ = $$\frac { 1 }{ r }$$, length of arc = l = 22 cm, radius = r = 100 cm, angle made at the centre = θ = ?
Applying formula θ = $$\frac { 1 }{ r }$$

Question 3.
In a circle of diameter 40 cm, the length of chord is 20 cm, then find the length of minor arc of the chord.
Solution:
Radius of circle = OA = OB = $$\frac { 40 }{ 2 }$$ = 20 cm, AB = 20 cm

Hence OAB is an equilateral triangle.
∴ θ = $$\frac { 1 }{ r }$$
⇒ 60° = $$\frac { AB }{ 20 }$$
⇒ AB = 60° x 20 = 60 x $$\frac { π }{ 180 }$$ x 20
= $$\frac { 20π }{ 3 }$$ cm.

Question 4.
If in two circles, arc of the same length substend angles of 60° and 75° at the centre, then find the ratio of their radii. (NCERT)
Solution:

Question 5.
Find value : (i) sin75°, (ii) tan15°
Solution:

Question 6.
Prove that:
sin2$$\frac { π }{ 4 }$$ + cos2$$\frac { π }{ 3 }$$ = – $$\frac { 1 }{ 2 }$$
Solution:

Question 7.
Prove that:
cos 27°. tan 27°. tan 63°. cosec 63° = 1. (NCERT)
Solution:
L.H.S. = cos 27°. tan 27°.tan 63°. cosec 63°
= cos 27°. tan 27° tan(90° – 27°). cosec (90° – 27°)
= cos 27°. tan 27° cot 27°. sec 27°
= $$\frac { 1 }{ sec 27° }$$.$$\frac { 1 }{ cot 27° }$$cot 27°. sec 27° sec 27° cot 27°
= 1 = R.H.S.

Question 8.
Find the general solution of the following equations :

1. sinθ = $$\frac { \sqrt { 3 } }{ 2 }$$
2. tanθ = 1
3. $$\sqrt { 3 }$$ tanθ + 1 = 0.

Solution:
1. sinθ = $$\frac { \sqrt { 3 } }{ 2 }$$
⇒ sinθ = sin$$\frac { π }{ 3 }$$
∴ θ = nπ + ( – 1)n$$\frac { π }{ 3 }$$ [ ∵ sinθ = sinα ⇒ θ = nπ + (- 1)nα, where n ∈ I]

2. tanθ = 1
⇒ tanθ = tan$$\frac { π }{ 4 }$$
∴ θ = nπ + $$\frac { π }{ 4 }$$ [ ∵tanθ = tanα ⇒ θ = nπ + α, where n ∈ I]

3. $$\sqrt { 3 }$$ tanθ + 1 = 0.
⇒ tanθ = – $$\frac { 1 }{ \sqrt { 3 } }$$ = – tan$$\frac { π }{ 6 }$$
⇒ tanθ = tan (π – $$\frac { π }{ 6 }$$)
⇒ tanθ = tan$$\frac { 5π }{ 6 }$$
∴ θ = nπ + $$\frac { 5π }{ 4 }$$ [ ∵tanθ = tanα ⇒ θ = nπ + α, where n ∈ I]

Question 9.
Solve the following equations :
(i) 4 sin2 θ = 1
(ii) 3 tan2 θ = 1.
Solution:

Question 10.
Find the principal value of x :
tan x = $$\sqrt { 3 }$$
Solution:
tan x =$$\sqrt { 3 }$$
tan x = tan$$\frac { π }{ 3 }$$ = tan(π + $$\frac { π }{ 3}$$)
⇒ tan x = tan$$\frac { π }{ 3 }$$ = tan$$\frac { 4π }{ 3 }$$
⇒ x = $$\frac { π }{ 3 }$$ or $$\frac { 4π }{ 3 }$$
Principal value of x = $$\frac { π }{ 3 }$$
∴ tan x = $$\sqrt { 3 }$$ ⇒ tanx = tan$$\frac { π }{ 3 }$$
Hence, general solution of x = nπ + $$\frac { π }{ 3 }$$, [ ∵tanθ = tanα ⇒ θ = nπ + α, where n ∈ I]

Question 11.
Find the principal value of x :
sec x = 2.
Solution:

Trigonometric Functions Long Answer Type Questions

Question 1.
Prove that:
$$\frac { sinx – siny }{ cosx + cosy }$$ = tan$$\frac { x – y }{ 2 }$$
Solution:

Question 2.
Prove that:
$$\frac { sin3x + sinx }{ cos3x + cosx }$$ = tan2x.
Solution:

Question 3.
Prove that:
$$\frac { sinx – sin3x }{ { sin }^{ 2 }x – { cos }^{ 2 }x }$$ = 2 sinx.
Solution:

Question 4.
Prove that:
$$\frac { cos9x – cos5x }{ sin17x – sin3x }$$ = – $$\frac { sin2x}{ cos10x }$$
Solution:

Question 5.
Prove that:
$$\frac { sin5x + sin3x }{ cos5x + cos3x }$$ = tan4x.
Solution:

Question 6.
Prove that:
2sin2$$\frac { π }{ 6 }$$ + cosec2$$\frac { 7π }{ 6 }$$cos2$$\frac { π }{ 3 }$$ = $$\frac { 3 }{ 2 }$$
Solution:

Question 7.
Prove that:
cot2$$\frac { π }{ 6 }$$ + cosec$$\frac { 5π }{ 6 }$$ + 3tan2$$\frac { π }{ 6 }$$ = 6
Solution:

Question 8.
Prove that:
2sin2$$\frac { 3π }{ 4 }$$ + 2cos2$$\frac { π }{ 4 }$$ + 2sec2$$\frac { π }{ 3 }$$ = 10
Solution:

Question 9.
Prove that:
$$\frac { cos11° + sin11° }{ cos11° – sin11° }$$ = tan56°
Solution:

Question 10.
Prove that:
tan 3A – tan 2A – tan A = tan 3A. tan 2A. tan A.
Solution:
tan3A – tan2A – tanA – tan3A.tan2A. tanA = 0
⇒ tan3A – tan2A – tanA (l + tan3A.tan2A) = 0
⇒ tan 3A – tan 2A = tan A (1 + tan 3 A. tan 2 A)
⇒ $$\frac { tan 3A – tan 2A}{ 1 + tan 2A.tan 3A }$$
⇒ tan(3A – 2 A) = tan A
⇒ tanA = tan A.

Question 11.
Prove that:
cos($$\frac { π }{ 4 }$$ + x) + cos($$\frac { π }{ 4 }$$ – x) = $$\sqrt {2}$$cosx
Solution:

Question 12.
Prove that:

Solution:

Question 13.
If α + β = $$\frac { π }{ 4 }$$, then prove that :
(1 + tanα)(1 + tanβ) = 2.
Solution:
Given α + β = $$\frac { π }{ 4 }$$
tan(α + β) = tan $$\frac { π }{ 4 }$$

⇒ tan α + tan β = 1 – tanα.tan β
⇒ tan α + tan β.tanα.tan β = 1
⇒ tan α + tan β(1 + tanα) = 1
⇒ 1 + tan α + tan β(1 + tanα) = 1 + 1
⇒ ( 1 + tan α)(1 + tan β) = 2

Question 14.
Find the general solution of cos4x = cos 2x.
Solution:
Given equation :
cos4x = cos2x
⇒ cos4x – cos2x = 0
⇒ – 2sin$$\frac { 4x + 2x}{ 2 }$$.sin$$\frac { 4x – 2x}{ 2 }$$ = 0
⇒ – 2sin3x.sinx = 0
⇒ sinx.sin3x = 0
⇒ sin3x = 0 or sinx = 0
⇒ 3x = nπ or x = nπ
⇒ x = $$\frac { nπ}{ 3 }$$ or x = nπ, where n∈I.

Question 15.
Solve the equation tan 2x = cot(x + $$\frac { π }{ 3 }$$)
Solution:

Question 16.
Solve the equation sin 3θ = sin 2θ
Solution:
Given equation sin3θ = sin2θ
sin 3θ – sin 2θ = 0

Question 17.
Solve the equation tan2θ = tan$$\frac { 2 }{ θ }$$
Solution:
tan 2θ = tan$$\frac { 2 }{ θ }$$
⇒ 2θ = nπ + $$\frac { 2 }{ θ }$$
⇒ 2θ – $$\frac { 2 }{ θ }$$ = nπ + $$\frac { 2 }{ θ }$$
⇒ 2θ – $$\frac { 2 }{ θ }$$ = nπ
⇒ 2θ2 – 2 = nπθ
⇒ 2θ2 – nπθ – 2 = 0

Question 18.
Solve the equation tanθ.tan2θ = 1
Solution:

Question 19.
If sec x = $$\frac { 13 }{ 5 }$$ and x is in fourth quadrant, then find the other five trigonometrical functions (NCERT)
Solution:
secx = $$\frac { 13 }{ 5 }$$ ,
Given : x is in fourth quadrant,

Question 20.
If tan x = – $$\frac { 5 }{ 12 }$$ and x is in second quadrant, then find the other five trigonometrical functions (NCERT)
Solution:
Given : tan x = – $$\frac { 5 }{ 12 }$$ ,

∴ x is in second quadrant, ∵ secx is negative.

∵ x is in second quadrant, hence sinx is positive.
sinx = $$\frac { 5 }{ 13 }$$
cosecx = $$\frac { 1 }{ sin x }$$ = $$\frac { 1 }{ 5/13 }$$ = $$\frac { 13 }{ 5 }$$

Question 21.
Prove that:
cos 20°. cos 40°. cos 60°. cos 80°= $$\frac { 1 }{ 16 }$$ .
Solution:
L.H.S. = cos 20° .cos 40°. cos 60°. cos 80°
= cos 20°.cos 40°.$$\frac { 1 }{ 2 }$$. cos80°
= $$\frac { 1 }{ 4 }$$(2 cos 20°. cos 40°). cos 80°
= $$\frac { 1 }{ 4 }$$(cos 60° + cos 20°). cos 80°
= $$\frac { 1 }{ 4 }$$ cos 60°. cos 80° + $$\frac { 1 }{ 4 }$$ cos 20°. cos 80°
= $$\frac { 1 }{ 4 }$$. $$\frac { 1 }{ 2}$$. cos 80° + $$\frac { 1 }{ 8 }$$(2 cos 20°. cos 80°)
= $$\frac { 1 }{ 8 }$$cos 80° + $$\frac { 1 }{ 8 }$$ (cos 100° + cos 60°)
= $$\frac { 1 }{ 8 }$$cos 80°+$$\frac { 1 }{ 8 }$$[cos(180° – 80°) + $$\frac { 1 }{ 2 }$$
= $$\frac { 1 }{ 8 }$$cos80° + $$\frac { 1 }{ 8 }$$(- cos80°) + $$\frac { 1 }{ 16 }$$ [∵cos(180° – θ]= – cosθ]
= $$\frac { 1 }{ 8 }$$cos 80° – $$\frac { 1 }{ 8 }$$cos 80°+ $$\frac { 1 }{ 16 }$$
= $$\frac { 1 }{ 16 }$$ = R.H.S.

Question 22.
Prove that:
sin 20°.sin40°.sin60°.sin80°=$$\frac { 3 }{ 16 }$$.
Solution:

Question 23.
prove that:
(cosx + cosy)2+ (sinx – siny)2 = 4cos2$$\frac { x + y }{ 2 }$$
Solution:

Question 24.
prove that:
(cosx – cosy)2+ (sinx – siny)2 = 4sin2$$\frac { x – y }{ 2 }$$
Solution:
L.H.S. = (cosx – cosy)2+ (sinx – siny)2
= cos2x + cos2y – 2 cosxcosy + sin2x + sin2y – 2sinx siny
= cos2x + sin2x + cos2y + sin2y – 2[cosx cosy + sinx siny]
= 1 + 1 – 2cos(x – y) = 2 – 2cos(x – y)
= 2[1 – cos(x – y)] = 2 x 2sin2
= 4 sin2$$\frac { x – y }{ 2 }$$ = R.H.S.

Question 25.
Prove that:
sinx + sin3x + sin5x + sin7x = 4 cosx cos2x sin4x.
Solution:
L.H.S. = sinx + sin3x + sin5x + sin 7x
= sin7x + sinx + sin5x + sin3x
= 2sin$$\frac { 7x + x }{ 2 }$$.cos$$\frac { 7x – x }{ 2 }$$ + 2sin$$\frac { 5x + 3x }{ 2 }$$.cos$$\frac { 5x – 3x }{ 2 }$$
= 2sin4x.cos3x + 2sin4x.cosx
= 2sin4x[cos3x + cosx]
= 2sin4x[ 2 x cos$$\frac { 3x + x }{ 2 }$$.cos$$\frac { 3x – x }{ 2 }$$ ]
= 2sin4x[2cos2x.cosx]
= 4sin4x.cos2x.cosx
= R.H.S

Question 26.
Prove that:

Solution:

Question 27.
Solve the equation and find general solution of sec22x = 1 – tan2x. (NCERT)
Solution:
The given equation is :
sec2 2x = 1 – tan 2x
⇒ 1 + tan22x = 1 – tan 2x
⇒ tan2 2x = – tan2x
⇒ tan2 2x + tan2x = 0 ⇒ tan2x (tan2x + 1) = 0
⇒ tan2x = 0 tan2x(tan2x + 1) = 0 ⇒ 2x = nπ or tan2x = – 1

Question 28.
Solve the equation 2 cos2 x + 3 sin x = 0
Solution:
2 cos2 x + 3 sin x = 0
⇒ 2 (1 – sinx)+3 sinx = 0
⇒ 2 – 2sin2 x+ 3 sinx = 0
⇒ 2 sin2 x – 3 sinx – 2 = 0
⇒ 2 sin2 x + sinx – 4sinx – 2 = 0
⇒ sinx(2 sinx + 1) – 2(2sinx + 1) = 0
⇒ (2sinx + 1)(sinx – 2) = 0
⇒ 2 sin x +1 = 0 or sinx – 2 = 0
⇒ 2 sin x = – 1 = 0 or sinx = 2
⇒ 2 sin x = – $$\frac { 1 }{ 2 }$$
⇒ sin x = sin(π + $$\frac { π }{ 6 }$$
⇒ sin x = sin$$\frac { 7π }{ 6 }$$
∴ x = nπ + (- 1)n$$\frac { 7π }{ 6 }$$

Question 29.
Solve the equation tan2 θ + (1 – $$\sqrt { 3 }$$) tanθ = $$\sqrt { 3 }$$
Solution:
tan2 θ + (1 – $$\sqrt { 3 }$$) tanθ = $$\sqrt { 3 }$$
⇒ tan2 θ + (1 – $$\sqrt { 3 }$$) tanθ – $$\sqrt { 3 }$$ = 0
⇒ tan2 θ + tanθ – $$\sqrt { 3 }$$ tanθ – $$\sqrt { 3 }$$ = 0
⇒ tanθ(tanθ + 1) – $$\sqrt { 3 }$$(tanθ + 1) = 0
⇒ (tanθ + 1)(tan θ – $$\sqrt { 3 }$$) = 0
⇒ tanθ + 1 = 0 or tan θ – $$\sqrt { 3 }$$ = 0
⇒ tanθ = – 1 or tan θ = $$\sqrt { 3 }$$
⇒ tanθ = tan $$\frac { – π }{ 4 }$$ = tanθ = tan$$\sqrt { 3 }$$
⇒ tanθ = tan $$\frac { – π }{ 4 }$$ = θ = nπ + $$\frac { π }{ 3 }$$
∴ θ = nπ – $$\frac { π }{ 4}$$

Question 30.
Solve the equation $$\sqrt { 2 }$$ secθ + tanθ = 1.
Solution:

Question 31.
Find the general solution of the equation sinx + sin5x + sin5x = 0
Solution:

Question 32.
Prove that:
2cos$$\frac { π }{ 13 }$$cos$$\frac { 9π }{ 13 }$$ + cos$$\frac { 3π }{ 13 }$$ + cos$$\frac { 5π }{ 3 }$$ = 0
Solution:

Question 33.
Prove that:

Solution: