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प्रश्न
Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]
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उत्तर
\[LHS = \left| \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right| + \left| \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right|\]
\[ = \left| \sqrt{\frac{\left( 1 - \sin x \right)\left( 1 - \sin x \right)}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)}} \right| + \left| \sqrt{\frac{\left( 1 + \sin x \right)\left( 1 + \sin x \right)}{\left( 1 - \sin x \right)\left( 1 + \sin x \right)}} \right|\]
\[ = \left| \sqrt{\frac{\left( 1 - \sin x \right)\left( 1 - \sin x \right)}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)}} \right| + \left| \sqrt{\frac{\left( 1 + \sin x \right)\left( 1 + \sin x \right)}{\left( 1 - \sin x \right)\left( 1 + \sin x \right)}} \right|\]
\[ = \left| \sqrt{\frac{\left( 1 - \sin x \right)^2}{1 - \sin^2 x}} \right| + \left| \sqrt{\frac{\left( 1 + \sin x \right)^2}{1 - \sin^2 x}} \right|\]
\[ = \left| \sqrt{\frac{\left( 1 - \sin x \right)^2}{\cos^2 x}} \right| + \left| \sqrt{\frac{\left( 1 + \sin x \right)^2}{\cos^2 x}} \right|\]
\[ = \left| \frac{1 - \sin x}{\cos x} \right| + \left| \frac{1 + \sin x}{\cos x} \right|\]
\[ = \left| \frac{1 - \sin x + 1 + \sin x}{\cos x} \right|\]
\[ = \left| \frac{2}{\cos x} \right|\]
\[ = - \frac{2}{\cos x} \left[ \because \frac{\pi}{2} < x < \pi \text{ and in the second quadrant, }\cos x \text{ is negative }\right]\]
= RHS
Hence proved .
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