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प्रश्न
Differentiate \[\sqrt{\frac{1 + \sin x}{1 - \sin x}}\] ?
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उत्तर
\[\text{Let } y = \sqrt{\frac{1 + \sin x}{1 - \sin x}}\]
\[\text{ Differentiate it with respect to x we get }, \]
\[\frac{d y}{d x} = \frac{d}{dx} \left( \frac{1 + \sin x}{1 - \sin x} \right)^\frac{1}{2} \]
\[ = \frac{1}{2} \left( \frac{1 + \sin x}{1 - \sin x} \right)^{\frac{1}{2} - 1} \frac{d}{dx}\left( \frac{1 + \sin x}{1 - \sin x} \right)\]
\[ = \frac{1}{2} \left( \frac{1 - \sin x}{1 + \sin x} \right)^\frac{1}{2} \left[ \frac{\left( 1 - \sin x \right)\left( \cos x \right) - \left( 1 + \sin x \right)\left( - \cos x \right)}{\left( 1 - \sin x \right)^2} \right]\]
\[ = \frac{1}{2}\frac{\left( 1 - \sin x \right)^\frac{1}{2}}{\left( 1 + \sin x \right)^\frac{1}{2}}\left[ \frac{\cos x - \cos x \sin x + \cos x + \sin x \cos x}{\left( 1 - \sin x \right)^2} \right]\]
\[ = \frac{1}{2} \times \frac{2\cos x}{\sqrt{1 + \sin x}\left( 1 - \sin x \right)\frac{3}{2}}\]
\[ = \frac{\cos x}{\sqrt{1 + \sin x}\left( 1 - \sin x \right)\frac{3}{2}}\]
\[ = \frac{\cos x}{\sqrt{1 + \sin x}\sqrt{1 - \sin x}\left( 1 - \sin x \right)}\]
\[ = \frac{\cos x}{\sqrt{1 - \sin^2 x} \times \left( 1 - \sin x \right)}\]
\[ = \frac{\cos x}{\cos x\left( 1 - \sin x \right)} \left[ \text{Using } 1 - \sin^2 x = \cos^2 x \right]\]
\[ = \frac{1}{\left( 1 - \sin x \right)} \times \frac{\left( 1 + \sin x \right)}{\left( 1 + \sin x \right)}\]
\[ = \frac{\left( 1 + \sin x \right)}{\left( 1 - \sin^2 x \right)}\]
\[ = \frac{1 + \sin x}{\cos^2 x}\]
\[ = \frac{1}{\cos x}\left( \frac{1}{\cos x} + \frac{\sin x}{\cos x} \right)\]
\[ = \sec x\left( \sec x + \tan x \right)\]
\[\text{Hence }, \frac{dy}{dx} = \sec x\left( \sec x + \tan x \right)\]
