Students can download Maths Chapter 6 Trigonometry Ex 6.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Question 1.
If sin 30° = x and cos 60° = y, then x² + y² is …….
(a) $$\frac{1}{2}$$
(b) 0
(c) sin 90°
(d) cos 90°
Solution:
(a) $$\frac{1}{2}$$
Hint:
sin 30° = x = $$\frac{1}{2}$$
cos 60° = y = $$\frac{1}{2}$$
x² + y²

Question 2.
If tan θ = cot 37°, then the value of θ is ………
(a) 37°
(b) 53°
(c) 90°
(d) 1°
Solution:
(b) 53°
Hint:
tan θ = cot 37°
= cot (90° – 53°)
= tan 53°
The value of θ is 53°

Question 3.
The value of tan 72°. tan 18° is ………
(a) 0
(b) 1
(c) 18°
(d) 72°
Solution:
(b) 1
Hint:
tan 72° . tan 18° = tan 72° . tan (90° – 72°)
= tan 72° . cot 72°
= tan 72° × $$\frac{1}{tan 72°}$$
= 1

Question 4.
The value of $$\frac{2 tan 30°}{1 – tan^{2} 30°}$$ is equal to ………
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Solution:
(c) tan 60°
Hint:

= √3 = tan 60°

Question 5.
If 2 sin 2θ = √3 then the value of θ is ………
(a) 90°
(b) 30°
(c) 45°
(d) 60°
Solution:
(b) 30°
Hint:
2 sin 2θ = √3 ⇒ sin 2θ = $$\frac{√3}{2}$$
sin 2θ = sin 60° ⇒ 2θ = 60°
θ = 30°

Question 6.
The value of 3 sin 70° sec 20° + 2 sin 39° sec 51° is ………
(a) 2
(b) 3
(c) 5
(d) 6
Solution:
(c) 5
Hint:
3 sin 70° sec 20° + 2 sin 39° sec 51°
= 3. sin 70° . sec (90° – 70°) + 2 sin 39° . sec (90° – 39°)
= 3. sin 70° . cosec 70° + 2 sin 39° . cosec 39°
= 3. sin70° × $$\frac{1}{sin 70°}$$ + 2 sin 39° × $$\frac{1}{sin 39°}$$
= 3 + 2
= 5

Question 7.
The value of $$\frac{1-tan^{2}45°}{1+tan^{2}45°}$$ is ……..
(a) 2
(b) 1
(c) 0
(d) $$\frac{1}{2}$$
Solution:
(c) 0
Hint:
$$\frac{1-tan^{2}45°}{1+tan^{2}45°}$$ = $$\frac{1-1}{1+1}$$
= $$\frac{0}{2}$$ = 0

Question 8.
The value of cosec (70° + θ) – sec (20° + θ) + tan (65° + θ) – cot (25° + θ) is ……..
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
cosec (70° + θ) – sec (20° – θ) + tan (65° + θ) – cot (25° – θ)
= sec [90° – (70° + θ)] – sec (20° – θ) + tan (65° + θ) – tan [90° – (25° – θ)]
= sec (20° – θ) – sec (20° – θ) + tan (65° + θ) – tan (65° + θ)
= 0

Question 9.
The value of tan 1° . tan 2° . tan 3°…. tan 89° is ………
(a) 0
(b) 1
(c) 2
(d) $$\frac{√3}{2}$$
Solution:
(b) 1
Hint:
tan 1° . tan 2° . tan 3° …….. tan 89°
= tan (90° – 89°). tan (90° – 88°) .tan (90° – 87°) …….. tan 45° . tan (89°)
= cot 89° . cot 88°. cot 87°. ……. tan 45° …….. tan 87°. tan 88°. tan 89°
= 1

Question 10.
Given that sin α = $$\frac{1}{2}$$ and cos β = $$\frac{1}{2}$$, then the value of α + β is ………
(a) 0°
(b) 90°
(c) 30°
(d) 60°
Solution:
Hint:
sin α = $$\frac{1}{2}$$
sin 30° = $$\frac{1}{2}$$ ∴ α = 30°
cos β = $$\frac{1}{2}$$ ∴ β = 60°
∴ α + β = 30° + 60°
= 90°