Question

The set of all possible values of $\alpha$ in $[-\pi ,\pi ]$ such that $\sqrt{\frac{1-\sin \alpha }{1+\sin \alpha }}$ is equal to $\sec \alpha -\tan \alpha$ is

• A. $[0,\frac{\pi }{2}]$
• B. $[0,\frac{\pi }{2}]\cup [\frac{\pi }{2},\pi ]$
• C. $[-\pi ,0]$
• D. $(-\frac{\pi }{2},\frac{\pi }{2})$

Answer 1

The correct option is D $(-\frac{\pi }{2},\frac{\pi }{2})$

suppose $y=\sqrt{\frac{1-sin\alpha }{1+sin\alpha }}=\sqrt{\frac{(1-sin\alpha {)}^2}{(1+sin\alpha )(1-sin\alpha )}}=\mid \frac{1-sin\alpha }{cos\alpha }\mid =\mid sec\alpha -tan\alpha \mid$

Given $y=sec\alpha -tan\alpha$

that means, $sec\alpha -tan\alpha >0$

$\Rightarrow (1-sin\alpha )/cos\alpha >0\Rightarrow cos\alpha >0$
therefore solution of $\alpha$ is,

$\alpha \in (\frac{-\pi }{2},\frac{\pi }{2})$

Hence, option 'D' is correct.