CBSE Class 10 Mathematics Basic 2024: question 30

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. Write $\cot \theta =\frac{1}{\tan \theta }$: LHS $=\frac{\tan \theta }{1-\frac{1}{\tan \theta }}+\frac{\frac{1}{\tan \theta }}{1-\tan \theta }=\frac{{\tan }^2\theta }{\tan \theta -1}+\frac{1}{\tan \theta (1-\tan \theta )}$

   ½ mark

2. $=\frac{{\tan }^2\theta }{\tan \theta -1}-\frac{1}{\tan \theta (\tan \theta -1)}=\frac{{\tan }^3\theta -1}{\tan \theta (\tan \theta -1)}$

   ½ mark

3. Using $a^3-b^3=(a-b)(a^2+ab+b^2)$: $=\frac{(\tan \theta -1)({\tan }^2\theta +\tan \theta +1)}{\tan \theta (\tan \theta -1)}$

   ½ mark

4. Cancel $(\tan \theta -1)$ and use $1+{\tan }^2\theta ={\sec }^2\theta$: $=\frac{{\sec }^2\theta +\tan \theta }{\tan \theta }$

   ½ mark

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

   ½ mark

6. $\frac{{\sec }^2\theta }{\tan \theta }=\frac{1}{{\cos }^2\theta }\cdot \frac{\cos \theta }{\sin \theta }=\frac{1}{\sin \theta \cos \theta }=\sec \theta \text{cosec}\theta$. So LHS $=1+\sec \theta \text{cosec}\theta =$ RHS.

   ½ mark

Answer: LHS = $1+\sec \theta \text{cosec}\theta$ = RHS (proved)

Revise the chapter

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