CBSE Class 10 Mathematics Standard 2026: question 28(a)

Hints

Open one hint at a time and try again before the next.

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. LHS $=\frac{{\sec }^3\theta }{{\tan }^2\theta }+\frac{{\mathrm{cosec}}^3\theta }{{\cot }^2\theta }$ [since ${\sec }^2\theta -1={\tan }^2\theta$, ${\mathrm{cosec}}^2\theta -1={\cot }^2\theta$]

   1 mark

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

   ½ mark

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

   ½ mark

4. $=\frac{1}{\sin \theta \cos \theta }\left[\frac{1}{\sin \theta }+\frac{1}{\cos \theta }\right]$

   ½ mark

5. $=\sec \theta \cdot \mathrm{cosec}\theta (\mathrm{cosec}\theta +\sec \theta )=$ RHS

   ½ mark

Answer: Proved: LHS = $\sec \theta \mathrm{cosec}\theta (\sec \theta +\mathrm{cosec}\theta )$

Revise the chapter

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