# Trigonometry Extra Question 13

#### Introduction to Trigonometry | Trigonometric Ratios of Specific Values

This class 10 Maths extra question is from chapter 8 of NCERT text book - Introduction to Trigonometry. Concept Tested: Trigonometric values of basic angles.

Question 13: Find the value of x in each of the following.
(i) cosec 3x = $$frac{\text{cot 30° + cot 60°}}{\text{1 + cot 30° cot ⁡60°}}$$ii) cos x = 2 sin 45° cos 45° – sin 30°

## NCERT Solution to Class 10 Maths

### Explanatory Answer | CBSE Extra Question 13

#### (i) cosec 3x = $$frac{\text{cot 30° + cot 60°}}{\text{1 + cot 30° cot ⁡60°}}$ cot 30° = √3 ; cot 60° = $\frac{1}{\sqrt{3}}$ $\frac{\text{cot ⁡30° + cot ⁡60° }}{\text{1 + cot 30° cot ⁡60° }}$ = $\frac{\sqrt{3}+ \frac{1}{\sqrt{3}}}{1 + \sqrt{3} × \frac{1}{\sqrt{3}}}$ = $\frac{\frac{$3+1$}{$sqrt3}}{2}) = $\frac{4}{\sqrt{3}}$ × $\frac{\text{1}}{\text{2}}$= $\frac{\text{2}}{\sqrt{3}}$ So, cosec 3x = $\frac{2}{\sqrt{3}}$ We know $\frac{\text{2}}{\sqrt{3}}$ = cosec 60° cosec 3x = cosec 60° ∴ 3x = 60° or x = 20° ####$ii) cos x = 2 sin 45° cos 45° – sin 30°

sin 45° = cos 45°= $\frac{\text{1}}{\sqrt{2}}$ ; sin 30° = $\frac{\text{1}}{\text{2}}$
cos x = 2 sin 45° cos 45° – sin 30°= [ 2 × $\frac{\text{1}}{\sqrt{2}}$ × $\frac{\text{1}}{\sqrt{2}}$ ]- $\frac{1}{2}$
cos x = 1 - $\frac{\text{1}}{\text{2}}$ = $\frac{\text{1}}{\text{2}}$
We know cos 60° = $\frac{\text{1}}{\text{2}}$
∴ cos x = cos 60° or x = 60°

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