Question
If $$y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 + \sin \,x} + \sqrt {1 - \sin \,x} }}{{\sqrt {1 + \sin \,x} - \sqrt {1 - \sin \,x} }}} \right]$$ where $$0 < x < \frac{x}{2},$$ then $$\frac{{dy}}{{dx}}$$ is equal to :
A.
$$\frac{1}{2}$$
B.
$$2$$
C.
$$\sin \,x + \cos \,x$$
D.
$$\sin \,x - \cos \,x$$
Answer :
$$\frac{1}{2}$$
Solution :
$$\eqalign{
& y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 + \sin \,x} + \sqrt {1 - \sin \,x} }}{{\sqrt {1 + \sin \,x} - \sqrt {1 - \sin \,x} }}} \right] \cr
& y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} + 2\,\sin \frac{x}{2}\cos \frac{x}{2}} + \sqrt {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} - 2\,\sin \frac{x}{2}\cos \frac{x}{2}} }}{{\sqrt {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} + 2\,\sin \frac{x}{2}\cos \frac{x}{2}} - \sqrt {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} - 2\,\sin \frac{x}{2}\cos \frac{x}{2}} }}} \right] \cr
& y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}} + \sqrt {{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}} }}{{\sqrt {{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}} - \sqrt {{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}} }}} \right] \cr
& y = {\cot ^{ - 1}}\left[ {\frac{{\cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2}}}{{\cos \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2} + \sin \frac{x}{2}}}} \right] \cr
& y = {\cot ^{ - 1}}\left[ {\frac{{2\,\cos \frac{x}{2}}}{{2\,\sin \frac{x}{2}}}} \right] \cr
& y = {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right) \cr
& y = \frac{x}{2} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2} \cr} $$