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Question
If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] is equal to
Options
sec x − tan x
sec x + tan x
tan x − sec x
none of these
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Solution
tan x − sec x
\[\sqrt{\frac{1 - \sin x}{1 + \sin x}} \]
\[ = \sqrt{\frac{\left( 1 - \sin x \right)\left( 1 - \sin x \right)}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)}}\]
\[ = \sqrt{\frac{\left( 1 - \sin x \right)^2}{1 - \sin^2 x}}\]
\[ = \sqrt{\frac{\left( 1 - \sin x \right)^2}{\cos^2 x}}\]
\[ = \frac{\left( 1 - \sin x \right)}{- cos x} \left[\text{ as,} \frac{\pi}{2} < x < \frac{3\pi}{2},\text{ so }\cos\theta\text{ will be negative }\right]\]
\[ = - \left( sec x - \tan x \right) \]
\[ = - sec x + \tan x\]
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