question_answer

Prove that \[{{\tan }^{-1}}\left( \frac{\cos x}{1-\sin x} \right)=\left( \frac{\pi }{4}+\frac{x}{2} \right),\] \[x\in \left( \frac{-\,\pi }{2},\frac{\pi }{2} \right).\]

Answer:

We have \[={{\tan }^{-1}}\left\{ \frac{2\sin \left( \frac{\pi }{4}-\frac{x}{2} \right)\cos \left( \frac{\pi }{4}-\frac{x}{2} \right)}{2{{\sin }^{2}}\left( \frac{\pi }{4}-\frac{x}{2} \right)} \right\}\] \[={{\tan }^{-1}}\left\{ \cot \left( \frac{\pi }{4}-\frac{x}{2} \right) \right\}={{\tan }^{-1}}\left[ \tan \left\{ \frac{\pi }{2}-\left( \frac{\pi }{4}-\frac{x}{2} \right) \right\} \right]\] \[={{\tan }^{-1}}\left\{ \tan \left( \frac{\pi }{4}+\frac{x}{2} \right) \right\}=\left( \frac{\pi }{4}+\frac{x}{2} \right)=RHS.\] Hence, \[{{\tan }^{-1}}\left( \frac{\cos x}{1-\sin x} \right)=\left( \frac{\pi }{4}+\frac{x}{2} \right).\]