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प्रश्न
If \[y = \cot^{- 1} \left\{ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right\}\], show that \[\frac{dy}{dx}\] is independent of x. ?
उत्तर
\[\text{ Let, y } = co t^{- 1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right] . . . \left( i \right)\]
\[\text{We have }, \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\]
\[ = \frac{\left( \sqrt{1 + \sin x} + \sqrt{1 - \sin x} \right)^2}{\left( \sqrt{1 + \sin x} - \sqrt{1 - \sin x} \right)\left( \sqrt{1 + \sin x} + \sqrt{1 - \sin x} \right)}\]
\[ = \frac{\left( 1 + \sin x \right) + \left( 1 - \sin x \right) + 2\sqrt{\left( 1 - \sin x \right)\left( 1 + \sin x \right)}}{\left( 1 + \sin x \right) - \left( 1 - \sin x \right)}\]
\[ = \frac{2 + 2\sqrt{1 - \sin^2 x}}{2\sin x}\]
\[ = \frac{1 + \cos x}{\sin x}\]
\[ = \frac{2 \cos^2 \frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\]
\[ = cot\frac{x}{2}\]
\[\text{ Therefore, equation } \left( i \right) \text{ becomes}\]
\[ y = co t^{- 1} \left( cot\frac{x}{2} \right)\]
\[ \Rightarrow y = \frac{x}{2}\]
\[ \therefore \frac{d y}{d x} = \frac{1}{2}\]
\[\text{ Hence }, \frac{d y}{d x} \text{ is independent of x} .\]
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