Question

Find the value of $\cos (\frac{3\pi }{2}+x)\cos (2\pi +x)\left[\cot \right(\frac{3\pi }{2}-x)+\cot (2\pi +x\left)\right]$.
Given, ${\sin }^{2}x+{\cos }^{2}x=1$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $1$

We know,
$\cos (\frac{3\pi }{2}+x)=\sin x\cos (2\pi +x)=\cos x$

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

$\therefore \cos (\frac{3\pi }{2}+x)\cos (2\pi +x)\times \left[\cot \right(\frac{3\pi }{2}-x)+\cot (2\pi +x\left)\right]$
$=\sin x.\cos x[\tan x+\cot x]$

$=\sin x.\cos x[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}]$

$=\sin x.\cos x\left[\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x.\cos x}\right]$

$=\sin x.\cos x\left[\frac{1}{\sin x.\cos x}\right]$
$=1$