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प्रश्न
Prove that: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
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उत्तर
\[LHS = \frac{\cos x}{1 - \sin x}\]
\[= \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} - 2\sin\frac{x}{2} \times \cos\frac{x}{2}} \left[ \because \text{ cos } x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}, \text{ sin } x = 2\sin\frac{x}{2}\cos\frac{x}{2} \text{ and } \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} = 1 \right]\]
\[ = \frac{\left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)}{\left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)^2}\]
\[ = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\]
On dividing the numerator and denominator by
\[\cos\frac{x}{2}\]
, we get
\[= \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\]
\[ = \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) = RHS\]
\[\text{ Hence proved } .\]
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