CBSE Class 10 Mathematics Basic 2023: question 29(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 $=\frac{1-\cos \theta }{1+\cos \theta }=\frac{(1-\cos \theta )(1-\cos \theta )}{(1+\cos \theta )(1-\cos \theta )}=\frac{(1-\cos \theta {)}^2}{1-{\cos }^2\theta }$

   1 mark

2. $=\frac{(1-\cos \theta {)}^2}{{\sin }^2\theta }$ [since $1-{\cos }^2\theta ={\sin }^2\theta$] $={\left(\frac{1-\cos \theta }{\sin \theta }\right)}^2$

   1 mark

3. $={\left(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }\right)}^2$

   ½ mark

4. $=(\mathrm{cosec}\theta -\cot \theta {)}^2=$ RHS. Hence proved.

   ½ mark

Answer: LHS = $(\mathrm{cosec}\theta -\cot \theta {)}^2$ = RHS

Revise the chapter

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