Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$

solution:

L.H.S. $=\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}$

$=\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}$

$=\frac{-\cos ^{2} x}{-\sin ^{2} x}$

$=\cot ^{2} x$

$=$ R.HS.