Question

Find the value of x:

$2{\tan }^{-1}\left(\sin x\right)={\tan }^{-1}\left(2{\sec }^{2}x\right),0<x<\frac{\pi }{2}$.

Answer 1

Given, $2{\tan }^{-1}\left(\sin x\right)={\tan }^{-1}\left(2{\sec }^{2}x\right),0<x<\frac{\pi }{2}$.

L.H.S.=$2{\tan }^{-1}\sin x$

We know that:

${\tan }^{-1}A+{\tan }^{-1}y={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

Therefore,

${\tan }^{-1}\sin x+{\tan }^{-1}\sin x={\tan }^{-1}\left(\frac{\sin x+\sin x}{1-{\sin }^{2}x}\right)$

$={\tan }^{-1}\left(\frac{2\sin x}{{\cos }^{2}x}\right)={\tan }^{-1}\left(2\sin x{\sec }^{2}x\right)$

Now,

${\tan }^{-1}\left(2\sin x{\sec }^{2}x\right)={\tan }^{-1}\left(2{\sec }^{2}x\right)$

Since, $0<x<\frac{\pi }{2}$

$2\sin x{\sec }^{2}x=2{\sec }^{2}x$

$\Rightarrow 2{\sec }^{2}x(\sin x-1)=0$

So, either $\sec x=0$, which is not possible

or $\sin x=1\Rightarrow x=\frac{\pi }{2}$