Q1. [2] § Introduction to Trigonometry (Chapter Header and Quote) § 8.5 Summary
If $\cot\theta = \frac{7}{8}$, then find the value of $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$.
Previously asked in CBSE board exam
2026 30/1/1 Q24(B)
Generated by claude-sonnet-4-6 · 2026-06-15 10:24 · grounding rag
Model Answer
Given: $\cot\theta = \dfrac{7}{8}$
$$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)} = \frac{1-\sin^2\theta}{1-\cos^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta$$
$$= \left(\frac{7}{8}\right)^2 = \boxed{\dfrac{49}{64}}$$
Source: Exercise 8.1, Q.7; Chapter 8 – Introduction to Trigonometry
---
Explanation
- The numerator $(1+\sin\theta)(1-\sin\theta) = 1-\sin^2\theta = \cos^2\theta$ and denominator $(1+\cos\theta)(1-\cos\theta) = 1-\cos^2\theta = \sin^2\theta$ — both use the identity $\sin^2\theta + \cos^2\theta = 1$.
- The ratio simplifies directly to $\cot^2\theta$, so you only need to square the given value. No need to find individual trig ratios.
- Examiners award 1 mark for the simplification step and 1 mark for the final value.
If a question refers to an image, map, graph or diagram that is not shown here, find the actual CBSE question paper on the CBSE website:
cbse.gov.in.