CBSE Class 10 Mathematics Standard sample paper 2026-27: question 23

Hints

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Step-by-step solution

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1. Given $\tan \theta =\frac{1}{\sqrt{5}}$, so $\cot \theta =\frac{1}{\tan \theta }=\sqrt{5}$.

2. Using ${\sec }^2\theta =1+{\tan }^2\theta$: ${\sec }^2\theta =1+\frac{1}{5}=\frac{6}{5}$.

   ½ mark

3. Using ${\mathrm{cosec}}^2\theta =1+{\cot }^2\theta$: ${\mathrm{cosec}}^2\theta =1+5=6$.

   ½ mark

4. Substitute: $\frac{{\mathrm{cosec}}^2\theta -{\sec }^2\theta }{{\mathrm{cosec}}^2\theta +{\sec }^2\theta }=\frac{6-\frac{6}{5}}{6+\frac{6}{5}}=\frac{\frac{24}{5}}{\frac{36}{5}}=\frac{24}{36}=\frac{2}{3}$.

   1 mark

Answer: $\frac{2}{3}$

Revise the chapter

• Chapter: Introduction to Trigonometry (notes, formulas, where marks are lost and every past question)
• Topic: Proving Simple Trigonometric Identities