Question

Evaluate : ${\tan }^{-1}\left(\cos ecx-\cot x\right)$

Answer 1

$T={\tan }^{-1}\left(\cos ecx-\cot x\right)={\tan }^{-1}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)={\tan }^{-1}\left(\frac{1-\cos x}{\sin x}\right)$

We know that

$1-\cos A=2{\sin }^2\left(\frac{A}{2}\right)$

$And,\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}$

Therefore,

$T={\tan }^{-1}\left\{\frac{\left(2{\sin }^2\frac{x}{2}\right)}{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}\right\}={\tan }^{-1}\left\{\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right\}={\tan }^{-1}\left\{\tan \frac{x}{2}\right\}=\frac{x}{2}$

$Whichistherequiredanswer.$