Proper way to solve $2\tan^{-1}(\cos x)=\tan^{-1}(2\csc x)$

Question

Solve

$$ 2\tan^{-1}(\cos x)=\tan^{-1}(2\csc x) $$

If $|\cos x|<1\implies x>0$, $$ \text{L.H.S}=\tan^{-1}\frac{2\cos x}{1-\cos^2x}=\tan^{-1}\frac{2\cos x}{\sin^2x}=\tan^{-1}\frac{2}{\sin x}\\ \implies {\sin x.\tan x}=\sin x\implies x=0 \text{ or }\frac{\pi}{4}\\ \boxed{x=\frac{\pi}{4}}\text{ as $x>0$} $$ If $|\cos x|>1\implies x<0$, $$ \text{L.H.S}=\pi+\tan^{-1}\frac{2\cos x}{1-\cos^2x}=\pi+\tan^{-1}\frac{2\cos x}{\sin^2x}=\tan^{-1}\frac{2}{\sin x}\\ $$

How do I prove second case does not hold in this problem ?

Answer 1

As $2\tan^{-1}(\cos x)=\tan^{-1}(2\csc x),$

$$-\dfrac\pi2<2\tan^{-1}(\cos x)<\dfrac\pi2$$

$$\iff-\tan\dfrac\pi4<\cos x<\tan\dfrac\pi4$$

as $\arctan y$ is an increasing function in real $y$

Answer 2

I'm not sure what benefit the division into cases can give.

Take the tangent of both sides; then, using the duplication formula for the tangent, you get $$ \frac{2\cos x}{1-\cos^2x}=\frac{2}{\sin x} $$ This simplifies to $$ \sin^2x=\sin x\cos x $$ This is $\sin x(\sin x-\cos x)=0$.

Since $\sin x\ne0$ (because of the cosecant in the statement), you remain with $$ \sin x-\cos x=0 $$