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

I have been trying to solve this since last night but still not able to solve it.Please help me to solve the above trigonometric functions. $$\cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi+x)\left[\cot\left(\frac{3\pi}{2}-x\right)+\cot(2\pi+x)\right]=1$$

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I'm not able to prove the above trigonometric function.Because of which i asked the question. – Drownpc Aug 29 '12 at 7:00

$\cos(3\frac{\pi}{2}+x)=\cos(2\pi+x-\frac{\pi}{2})=\cos(x-\frac{\pi}{2})=\cos(\frac{\pi}{2}-x)=\sin x$

$\cos(2\pi+x)=\cos x$

$\cot(3\frac{\pi}{2}-x)=\cot(2\pi-(\frac{\pi}{2}+x))=-\cot(\frac{\pi}{2}+x)=-(-\tan x)=\tan x$

$\cot(2\pi+x)=\cot x$

$LHS=(\sin x\cos x)(\tan x+\cot x)=(\sin x\cos x)(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})=\sin^2x+\cos^2x=1$ assuming $\sin x\cos x ≠ 0$

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Thank you for helping me to solve this question. – Drownpc Aug 29 '12 at 7:07

$\cos(3\pi/2+x)=\cos(\pi+x+pi/2)=-\cos(\pi/2+x)=+\sin(x)$
$\cos(2\pi+x)=\cos(x)$
$\cot((3\pi/2)-x)=\cot(\pi-(x-\pi/2))=-\cot(x-\pi/2)=\cot(\pi/2-x)=\tan(x)$
$\cot(2\pi+x)=\cot(x)$
So:
$$[\sin(x)\times \cos(x)]\times [\tan(x)+\cot(x)]=\sin^2(x)+\cos^2(x)=1$$

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