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

Choose the correct answer or most suitable answer:

Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 1
(a) \(\sqrt{2}\)
(b) \(\sqrt{3}\)
(c) 2
(d) 4
Solution:
(d) 4
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 20

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

Question 2.
If cos 28° + sin 28° = k3, then cos 17° is equal to …….
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 21
Solution:
(a) \(\frac{k^{3}}{\sqrt{2}}\)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 22

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 23
(a) 4 + \(\sqrt{2}\)
(b) 3 + \(\sqrt{2}\)
(c) 9
(d) 4
Solution:
(a) 4 + \(\sqrt{2}\)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 24

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 25
Solution:
(a) \(\frac{1}{8}\)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 26

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 27
Solution:
(c) 2 cos θ
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 28
Here θ is in II Quadrant

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

Question 6.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 29
Solution:
(d) \(\text { (d) } \frac{1-\lambda^{2}}{2 \lambda}\)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 30

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.

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

Question 8.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 31
Solution:
(b) \(\frac{1}{12}\)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 32

Question 9.
Which of the following is not true?
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 33
Solution:
(d) sec θ = \(\frac{1}{4}\)
Hint:
sec θ = \(\frac{1}{4}\) ⇒ cos θ = 4, which is not true.

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

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.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 60
(a) sin A + sin B + sin C
(b) 1
(c) 0
(d) cos A + cos B + cos C
Solution:
(c) 0
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 61
= tan A – tan B
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 62

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

Question 12.
cos pθ + cos qθ = 0 and if p ≠ q, then 0 is equal to (n is any integer) …….
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 63

Solution:
(b) \(\frac{\pi(2 n+1)}{p \pm q}\)
Hint:
cos pθ + cos qθ = 0
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 64

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
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 65

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

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.
On simplifying we get
sin2 A + sin2 B + sin2 C = 2 + 2 cos A cos B cos C
= 2 (given)
⇒ cos A cos B cos C = 0
cos A (or) cos B (or) cos C = 0
⇒ A (or) B (or) C = π/2
⇒ ABC (is a right angled 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:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 70
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 71

Question 16.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 73
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]
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 74

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

Question 17.
The triangle of maximum area with constant perimeter 12m
(a) is an equilateral triangle with side 4m
(b) is an isosceles triangle with sides 2m, 5m, 5m
(c) is a triangle with sides 3m, 4m, 5m
(d) does not exists
Solution:
(a) is an equilateral triangle with side 4m
Hint.
A triangle will have a max area (with a given perimeter) when it is an equilateral triangle.

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.
1 rotation makes 2πc
Distance travelled in 1 second = 2 radians
So time taken to complete 10 rotations = 6 × 2π = 20 πc
\(=\frac{20 \pi}{2}=10 \pi\) seconds

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

Question 19.
If sin α + cos α = b, then sin 2α is equal to ……..
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 80
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

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12

Question 20.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 81
(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.
When A + B + C = 180°
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12 82
When A + B + C = 180° each angle will be lesser than 180°
So sin A, sin B, sin C > 0
⇒ sin A sin B sin C > 0
So both (i) and (ii) are true

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