Question

Solve :
$\frac{1+\cos x}{\sin x\cos x}$

Answer 1

$\frac{1+\cos x}{\sin x\cos x}\Rightarrow \frac{2{\cos }^2\frac{x}{2}}{\sin x\cos x}$

$\Rightarrow \frac{2{\cos }^2\frac{x}{2}}{\left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)\left(\cos x\right)}$

$\Rightarrow \frac{\cos \frac{x}{2}}{\left(\sin \frac{x}{2}\right)\left(cosx\right)}$

$\Rightarrow \frac{\left(\cos \frac{x}{2}\right)}{\left(\sin \frac{x}{2}\right)(2{\cos }^2\frac{x}{2}-1)}$

$\Rightarrow \frac{\tan \frac{x}{2}}{(2{\cos }^2\frac{x}{2}-1)}$

$\Rightarrow \frac{\tan \frac{x}{2}}{(\frac{2}{{\sec }^2\frac{x}{2}}-1)}$

$\Rightarrow \frac{\left(\tan \frac{x}{2}\right)\left({\sec }^2\frac{x}{2}\right)}{(2-{\sec }^2\frac{x}{2})}$

$\Rightarrow \frac{\left(\tan \frac{x}{2}\right)(1+{\tan }^2\frac{x}{2})}{(1+(1-{\sec }^2\frac{x}{2}))}$

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