$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
Question.
$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
solution:
L.H.S. $=\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$
$=\sin x \cos x[\tan x+\cot x]$
$=\sin x \cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$
$=(\sin x \cos x)\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right]$
$=1=\mathrm{R} . \mathrm{H} \mathrm{S}$
L.H.S. $=\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$
$=\sin x \cos x[\tan x+\cot x]$
$=\sin x \cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$
$=(\sin x \cos x)\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right]$
$=1=\mathrm{R} . \mathrm{H} \mathrm{S}$
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