Question

Prove that: $\cos (\frac{3\pi }{2}+x)\cos (2\pi +x)[\cot (\frac{3\pi }{2}-x)+\cot (2\pi +x\left)\right]=1$

Answer 1

$L.H.S=\cos (\frac{3\pi }{2}+x)\cos (2\pi +x)[\cot (\frac{3\pi }{2}-x)+\cot (2\pi +x\left)\right]$
$=\sin x\times \cos x\times [\tan x+\cot x]$
$=\sin x\times \cos x\times [\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}]$
$=\sin x\times \cos x\times [\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}]$
$=\sin x\times \cos x\times \left[\frac{{\sin }^2+{\cos }^{2}x}{\sin x\cos x}\right]$
$={\sin }^{2}x+{\cos }^{2}x$
$=1=R.H.S$.
So, L.H.S. = R.H.S.
Hence proved.