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) cosec^{2} 45° + tan^{2} 45 ° – 3sin^{2} 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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cosec 45° = √2; tan 45° = 1; sin 90° = 1

∴ cosec^{2} 45° + tan^{2} 45° – 3 sin^{2} 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

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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CBSE Class 10 Maths - 2020

- Real Numbers Revision Class
- Polynomials Revision Videos
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- Arithmetic Progressions Revision Video
- Triangles Revision
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