STD 12 - 5. continuity and differentiation — JEE कक्षा 11 साइन्स — Question

$ = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}{{{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2}}}} \right)} \right]$

$ = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{1 - \tan \left( {\frac{x}{2}} \right)}}{{1 + \tan \left( {\frac{x}{2}} \right)}}} \right)} \right] = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right)} \right] = - \frac{1}{2}$