Extra Questions Class 10 Maths Chapter 8 | Q12

Trigonometry | Trigonometric Values of Basic Angles

This class 10 Maths extra question is from the chapter Introduction to Trigonometry. Concept: Solve trigonometric equations involving trigonometric values of basic angles.

Question 12 : Evaluate the following.
(i) cosec2 45° + tan2 45 ° – 3sin2 90°
(ii) cos 60° cos 30° – sin 60° sin 30°
(iii) $$frac{\text{tan 60° − tan 30°}}{\text{1 +tan 60° tan 30°}}$ Target Centum in CBSE 10th Maths Online CBSE Course online.maxtute.com Video Explanation NCERT Solution to Class 10 Maths With Videos Explanatory Answer | Trigonometry Extra Question 12 $i) cosec2 45° + tan2 45° – 3sin2 90°

cosec 45° = √2; tan 45° = 1; sin 90° = 1
∴ cosec2 45° + tan2 45° – 3 sin2 90° = (2 + 1) – (3 × 1) = 0

cos 60° =$$frac{1}{2}$; cos 30° =$\frac{\sqrt{3}}{2}$; sin 60° =$\frac{\sqrt{3}}{2}$ and sin 30° = $\frac{1}{2}$ ∴ cos 60° cos 30° – sin 60° sin 30° = [$\frac{1}{2}$ × $\frac{\sqrt{3}}{2}$] - [$\frac{\sqrt{3}}{2}$ × $\frac{1}{2}$] = 0 iii) $\frac{\text{tan 60° − tan 30°}}{\text{1 +tan 60° tan 30°}}$ tan 60° = √3; tan 30°= $\frac{1}{\sqrt{3}}$ $\frac{\text{tan 60° − tan 30° }}{\text{1 + tan ⁡60° tan 30°}}$ = $\frac{\sqrt{3} − \frac{1}{\sqrt{3}}}{1 +$$sqrt{3} × \frac{1}{\sqrt{3}}$}) = $\frac{\frac{$3 − 1$}{$sqrt{3}}}{1 + 1}$
= $\frac{2}{\sqrt{3}}$ × $\frac{1}{2}$= $\frac{1}{\sqrt{3}}$

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