Extra Questions For Class 10 Maths Chapter 8 | Q12

Trigonometry | Trigonometric Values of Basic Angles

This class 10 Maths extra Practice Question from the topic 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°}})


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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

(ii) cos 60° cos 30° – sin 60° sin 30°

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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