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

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. Take the LCM of the two square-root terms. The common denominator is $\sqrt{(\csc \theta +1)(\csc \theta -1)}$:
   $$\text{LHS}=\frac{(\csc \theta -1)+(\csc \theta +1)}{\sqrt{(\csc \theta +1)(\csc \theta -1)}}$$

   1 mark

2. $$=\frac{2\csc \theta }{\sqrt{{\csc }^2\theta -1}}$$

   1 mark

3. Since ${\csc }^2\theta -1={\cot }^2\theta$ (taking $\theta$ acute):
   $$=\frac{2\csc \theta }{\cot \theta }$$

   ½ mark

4. $$=2\cdot \frac{1}{\sin \theta }\cdot \frac{\sin \theta }{\cos \theta }=\frac{2}{\cos \theta }=2\sec \theta =\text{RHS}$$
   Hence proved.

   ½ mark

Answer: LHS = $\frac{2\csc \theta }{\cot \theta }=2\sec \theta$ = RHS

Revise the chapter

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