Question

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

Answer 1

$\frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos [\frac{\pi }{2}+x]}$

$=\frac{-\cos x\cdot \cos x}{\sin x(-\sin x)}=\frac{{\cos }^{2}x}{{\sin }^{2}x}={\cot }^{2}x$