CBSE Class 10 Mathematics Standard 2026: question 24(B)

Hints

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Step-by-step solution

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1. Multiply out the brackets using $(1+a)(1-a)=1-a^2$:
   $$\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\frac{1-{\sin }^2\theta }{1-{\cos }^2\theta }$$

   1 mark

2. Since ${\sin }^2\theta +{\cos }^2\theta =1$, we have $1-{\sin }^2\theta ={\cos }^2\theta$ and $1-{\cos }^2\theta ={\sin }^2\theta$. So the expression is
   $$\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta$$

   ½ mark

3. Put $\cot \theta =\frac{7}{8}$:
   $${\cot }^2\theta ={\left(\frac{7}{8}\right)}^2=\frac{49}{64}$$

   ½ mark

Answer: $\frac{49}{64}$

Revise the chapter

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