Question

If $\tan \left(\cot x\right)=\cot \left(\tan x\right)$, then $\sin 2x$ is equal to

• A. $\frac{2}{(2n+1)\pi }$
• B. $\frac{4}{(2n+1)\pi }$
• C. $\frac{2}{n(n+1)\pi }$
• D. $\frac{4}{n(n+1)\pi }$

Answer 1

Answer: (B)

$\frac{4}{(2n+1)\pi }$

Given, $\tan \left(\cot x\right)=\cot \left(\tan x\right)=\tan (\frac{\pi }{2}-\tan x)$

$\Rightarrow \cot x=n\pi +\frac{\pi }{2}-\tan x$

$\Rightarrow \cot x+\tan x=n\pi +\frac{\pi }{2}$

$\Rightarrow \frac{1}{\sin x\cos x}=n\pi +\frac{\pi }{2}$

$\Rightarrow \frac{1}{\sin 2x}=\frac{n\pi }{2}+\frac{\pi }{4}$

$\Rightarrow \sin 2x=\frac{1}{\frac{n\pi }{2}+\frac{\pi }{4}}=\frac{4}{(2n+1)\pi }$.