Question

The value of $\cos \left(\frac{3\pi }{2}+x\right)\cos (2\pi +x)\left\{\cot \left(\frac{3\pi }{2}-x\right)+\cot (2\pi +x)\right\}$ is

• A. $0$
• B. $1$
• C. $cosx$
• D. $sinx$

Answer 1

Answer: (B)

$\cos \left(\frac{3\pi }{2}+x\right)\cos (2\pi +x)$
$\left\{\cot \left(\frac{3\pi }{2}-x\right)+\cot (2\pi +x)\right\}$ $=\sin x.\cos x(\tan x+\cot x\}$
$=\sin x.\cos x\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x.\cos x}\right)$
$=1$
$(\because {\sin }^2\theta +{\cos }^2\theta =1)$