Mathematics — Standard (041) — CBSE Class 10 board question

LHS $= \sqrt{\dfrac{1-\sin\theta}{1+\sin\theta}}$

Multiply numerator and denominator inside the root by $(1-\sin\theta)$:

$$= \sqrt{\dfrac{(1-\sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)}} = \sqrt{\dfrac{(1-\sin\theta)^2}{1-\sin^2\theta}}$$

Since $1 - \sin^2\theta = \cos^2\theta$:

$$= \sqrt{\dfrac{(1-\sin\theta)^2}{\cos^2\theta}} = \dfrac{1-\sin\theta}{\cos\theta} = \dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta} = \sec\theta - \tan\theta = \textbf{RHS}$$

Hence proved.

Source: Chapter 8, Exercise 8.3 (trigonometric identities)

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• The standard technique is to rationalise the expression under the root by multiplying by $(1-\sin\theta)/(1-\sin\theta)$, converting the denominator to $\cos^2\theta$ using the identity $\sin^2\theta + \cos^2\theta = 1$.
• Then the square root simplifies cleanly to $\frac{1-\sin\theta}{\cos\theta}$, which splits into $\sec\theta - \tan\theta$.
• Examiners award marks for: correct rationalisation step, applying the identity, taking the square root, and splitting the fraction. Show each step clearly.