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°


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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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