CBSE Class 10 Mathematics Standard 2025: question 27(a)

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Since $\frac{1}{\tan \theta }=\cot \theta$ and $\frac{1}{\cot \theta }=\tan \theta$:
   $$\text{LHS}=(1+{\cot }^2\theta )(1+{\tan }^2\theta )$$

   1 mark

2. Using the identities $1+{\cot }^2\theta ={\csc }^2\theta$ and $1+{\tan }^2\theta ={\sec }^2\theta$:
   $$\text{LHS}={\csc }^2\theta \cdot {\sec }^2\theta$$

   ½ mark

3. $$=\frac{1}{{\sin }^2\theta }\cdot \frac{1}{{\cos }^2\theta }$$

   ½ mark

4. Using ${\cos }^2\theta =1-{\sin }^2\theta$:
   $$=\frac{1}{{\sin }^2\theta (1-{\sin }^2\theta )}$$

   ½ mark

5. $$=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }=\text{RHS}$$
   Hence proved.

   ½ mark

Answer: LHS = ${\csc }^2\theta {\sec }^2\theta =\frac{1}{{\sin }^2\theta -{\sin }^4\theta }$ = RHS

Revise the chapter

• Chapter: Introduction to Trigonometry (notes, formulas, where marks are lost and every past question)