# Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

Question 1. (a) $$\sqrt{2}$$
(b) $$\sqrt{3}$$
(c) 2
(d) 4
Solution:
(d) 4
Hint:  Question 2.
If cos 28° + sin 28° = k3, then cos 17° is equal to ……. Solution:
(a) $$\frac{k^{3}}{\sqrt{2}}$$
Hint: Question 3. (a) 4 + $$\sqrt{2}$$
(b) 3 + $$\sqrt{2}$$
(c) 9
(d) 4
Solution:
(a) 4 + $$\sqrt{2}$$
Hint: Question 4. Solution:
(a) $$\frac{1}{8}$$
Hint: Question 5. Solution:
(c) 2 cos θ
Hint: Here θ is in II Quadrant Question 6. Solution:
(d) $$\text { (d) } \frac{1-\lambda^{2}}{2 \lambda}$$
Hint: Question 7.
cos 1° + cos 2° + cos 3° + …. + cos 179° = …….
(a) 0
(b) 1
(c) -1
(d) 89
Solution:
(a) 0
Hint:
LHS = (cos 10 + cos 179°)+(cos 2° ÷ cos 178°)+ ….. +cos(89° + cos 91°)+cos 90°
cos 179° = cos (180° – 1) = – cos 1°
cos 178° = cos(180° – 2)= – cos 2°
So (cos 1°- cos 1°)+(cos 2° – cos 2°) + (cos 89° – cos 89°) + cos 90°
= 0 + 0 …. + 0 + 0 = 0. Question 8. Solution:
(b) $$\frac{1}{12}$$
Hint: Question 9.
Which of the following is not true? Solution:
(d) sec θ = $$\frac{1}{4}$$
Hint:
sec θ = $$\frac{1}{4}$$ ⇒ cos θ = 4, which is not true. Question 10.
cos 2θ cos 2ϕ+ sin2 (θ – ϕ) – sin2 (θ + ϕ) is equal to …….
(a) sin 2(θ + ϕ)
(b) cos 2(θ + ϕ)
(c) sin 2(θ – ϕ)
(d) cos 2(θ – ϕ)
Solution:
(b) cos 2(θ + ϕ)
Hint.
Given cos 2θ . cos 2ϕ + sin2 (θ – ϕ) – sin2 (θ + ϕ)
= cos 2θ cos 2ϕ + sin (θ – ϕ + θ + ϕ) sin (θ – ϕ – θ – ϕ)
= cos 2θ cos 2ϕ + sin 2θ sin(-2ϕ)
= cos 2θ cos 2ϕ – sin 2θ sin(2ϕ)
= cos (2θ + 2ϕ) = cos 2(θ + ϕ)

Question 11. (a) sin A + sin B + sin C
(b) 1
(c) 0
(d) cos A + cos B + cos C
Solution:
(c) 0
Hint: = tan A – tan B  Question 12.
cos pθ + cos qθ = 0 and if p ≠ q, then 0 is equal to (n is any integer) ……. Solution:
(b) $$\frac{\pi(2 n+1)}{p \pm q}$$
Hint:
cos pθ + cos qθ = 0 Question 13.
If tan α and tan β are the roots x2 + ax + b = 0, then $$\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}$$ is equal to …….
Solution:
(b) $$\frac{a}{b}$$
Hint:
tan2 x + a tan x + b = 0
α and β are the roots of the equation
⇒ tan2 α + a tan α + b = 0 ….. (1)
tan2 β + a tan β + b = 0 …… (2)
(1) – (2) => tan2 α – tan2 β + a (tan α – tan β) = 0
(tan α – tan β) (tan α + tan β) + a (tan α – tan β) = 0
⇒ tan α + tan β = – a …. (A)
(1) × tan β – (2) × tan α  Question 14.
In a triangle ABC, sin2 A + sin2 B + sin2 C = 2, then the triangle is …….
(a) equilateral triangle
(b) isosceles triangle
(c) right triangle
(d) scalene triangle
Solution:
(c) right triangle
Hint.
Given sin2 A + sin2 B + sin2 C = 2
Suppose the given triangle is a right angle triangle with ∠C = 90°, then
sin2 C = sin 2 90° = 1 …….. (1)
∴ sin2 A + sin2 B + 1 = 2
sin2 A + sin2B = 1
Also A + B = 90° ⇒ A = 90° – B
sin A = sin (90° – B) = cos B ——- (2)
sin2 A + sin2 B + sin2 C = 2
Using equations (1) and (2)
⇒ cos2 B + sin2 B + sin2 90° = 2
1 + 1 = 2
2 = 2
∴ sin2 A + sin2 B + sin2 C = 2 is true.
∴ ∆ ABC is a right-angle triangle.

Question 15.
If f(θ) = |sin θ| + |cos θ|, θ ∈ R, then f(θ) is in the interval …….
(a) [0, 2]
(b) [1, $$\sqrt{2}$$]
(c) [1, 2]
(d) [0, 1]
Solution:
(b) [1, $$\sqrt{2}$$]
Hint:  Question 16. Solution:
Nr: cos 6x + cos 4x + 5 cos 4x + 5 cos 2x + 10 cos 2x + 10
= 2 cos 5x cos x+5(2 cos 3x cos x)+ 10(2 cos2x)
= 2 cos x [cos 5x + 5 cos 3x + 10 cos x]  Question 17.
The triangle of maximum area with constant perimeter 12m
(a) is an equilateral triangle with side of 4m
(b) is an isosceles triangle with sides 2m, 5m, 5m
(c) is a triangle with sides 3m, 4m, 5m
(d) does not exist
Solution:
(a) is an equilateral triangle with side of 4m
Hint.
Given the perimeter of the triangle is 12m
2s = 12 ⇒ s = 6
Maximum area is obtained when it is an equilateral triangle with a side of 4m each.

Question 18.
A wheel is spinning at 2 radians/second. How many seconds will it take to make 10 complete rotations?
(a) 10π seconds
(b) 20π seconds
(c) 5π seconds
(d) 15π seconds
Solution:
(a) 10π seconds
Hint.
In 1 second, it rotates = 2 radians
For 2 radians rotation time taken = 1 second
∴ For 1 complete rotation (2 π radians) time taken
= $$\frac { 1 }{ 2 }$$ × 2π = π seconds.
∴ For 10 revolution time taken = π × 10
= 10 π seconds. Question 19.
If sin α + cos α = b, then sin 2α is equal to …….. Solution:
(b) b2 – 1, if b > $$\sqrt{2}$$
Hint:
sin α + cos β = b
(sin α + cos β)2 = b2
sin2 α + cos2 α + 2 sin α cos α = b2
sin2 α = b2 – 1 Question 20. (a) Both (i) and (ii) are true
(b) Only (i) is true
(c) Only (ii) is true
(d) Neither (i) nor (ii) is true
Solution:
(a) Both (i) and (ii) are true
Hint.
We know in a ∆ ABC
sin $$\frac{\mathbf{A}}{2}$$ sin $$\frac{\mathbf{B}}{2}$$ sin $$\frac{\mathbf{C}}{2}$$ > 0
Also sin A sin B sin C > 0
∴ Statements (i) and (ii) are true.