Question

If \[\frac{\pi}{2} < x < \pi, \text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}}\] is equal to

विकल्प

• 2 sec x

• −2 sec x

• sec x

• −sec x

Answer 1

−2 sec x

\[\sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}}\]
\[ = \sqrt{\frac{\left( 1 - \sin x \right)\left( 1 - \sin x \right)}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)}} + \sqrt{\frac{\left( 1 + \sin x \right)\left( 1 + \sin x \right)}{\left( 1 - \sin x \right)\left( 1 + \sin x \right)}}\]
\[ = \sqrt{\frac{\left( 1 - \sin x \right)^2}{1 - \sin^2 x}} + \sqrt{\frac{\left( 1 + \sin x \right)^2}{1 - \sin^2 x}}\]
\[ = \sqrt{\frac{\left( 1 - \sin x \right)^2}{\cos^2 x}} + \sqrt{\frac{\left( 1 + \sin x \right)^2}{\cos^2 x}}\]
\[ = \frac{\left( 1 - \sin x \right)}{- \cos x} + \frac{\left( 1 + \sin x \right)}{- \cos x} \left[ \frac{\pi}{2} < x < \pi, \text{so }\cos x \text{ will be negative . }\right]\]
\[ = - \left( \sec x - \tan x \right) - \left( \sec x + \tan x \right)\]
\[ = - 2\sec x\]