Question

If $\tan (\cot \, x) = \cot (\tan \, x)$, 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)

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

Answer 2

Given, $\tan\left(\cot x\right) = \cot\left(\tan x\right) = \tan \left(\frac{\pi}{2} -\tan x\right)$ $\Rightarrow \cot x = n \pi+ \frac{\pi}{2} - \tan x$ $\Rightarrow \cot x + \tan x = n \pi+ \frac{\pi}{2}$ $\Rightarrow \frac{1}{\sin x \cos x} = n \pi+ \frac{\pi}{2} \Rightarrow \frac{1}{\sin2x} = \frac{n\pi}{2} + \frac{\pi}{4}$ $\Rightarrow \sin2x = \frac{1}{\frac{n\pi}{2} + \frac{\pi}{4}} = \frac{4}{\left(2n+1\right)\pi}$