Question

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

Answer 1

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

$=\frac{-\cos x\left(\cos x\right)}{\sin x(-\sin x)}$

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

$={\cot }^{2}x$