Question
The set of all possible values of $$\alpha $$ in $$\left[ { - \pi ,\pi } \right]$$ such that $$\sqrt {\frac{{1 - \sin \alpha }}{{1 + \sin \alpha }}} $$ is equal to $$\sec \alpha - \tan \alpha $$ is
A.
$$\left[ {0,\frac{\pi }{2}} \right)$$
B.
$$\left[ {0,\frac{\pi }{2}} \right) \cup \left( {\frac{\pi }{2},\pi } \right)$$
C.
$$\left[ { - \pi ,0} \right]$$
D.
$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
Answer :
$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
Solution :
Clearly, $$\alpha \ne \pm \frac{\pi }{2}.$$
$$\eqalign{
& \sec\alpha - \tan\alpha = \frac{{1 - \sin \alpha }}{{\cos \alpha }}\,\,{\text{and}} \cr
& \sqrt {\frac{{1 - \sin \alpha }}{{1 + \sin \alpha }}} = \sqrt {\frac{{{{\left( {1 - \sin \alpha } \right)}^2}}}{{{{\cos }^2}\alpha }}} = \left| {\frac{{1 - \sin \alpha }}{{\cos \alpha }}} \right| = \frac{{1 - \sin \alpha }}{{\left| {\cos \alpha } \right|}}. \cr} $$
Hence, these will be equal if $$\cos\alpha > 0,\,{\text{i}}{\text{.e}}{\text{., }} - \frac{\pi }{2} < \alpha < \frac{\pi }{2}.$$