Proving $\arctan(\sinh(x))= 2\arctan(e^x) - \frac{\pi}{2}$

Question

I’m attempting to prove this:
$$ f(x) = \arctan(e^x) \\ g(x) = \arctan(\sinh(x)) \\ \forall x \in \mathbb{R} \\ g(x) = 2f(x) - \frac{\pi}{2} $$ Taking the tan of both sides, yields good results. However I'm unsure whether i can do that to the second part of this equation, as it lies within an interval larger than of $\left[\dfrac{-\pi}2,\dfrac{\pi}2\right]$.

Answer 1

You can always take the tangent of both sides, since the values (of the RHS, which is what you are concerned about) are bounded between $(-\pi/2,\pi/2)$ and both sides are monotonically increasing. You can see the boundedness as follows.

$\lim_{x \rightarrow -\infty} 2 \arctan(e^{x})-\frac{\pi}{2}=0-\frac{\pi}{2}=-\frac{\pi}{2}$

and

$\lim_{x \rightarrow \infty} 2 \arctan(e^{x})-\frac{\pi}{2}=2\frac{\pi}{2}-\frac{\pi}{2}=\frac{\pi}{2}$.

If you take $\tan$ on both sides, you get for the left hand side $\tan(\arctan(\sinh(x)))=\sinh(x)$ and for the right hand side $\tan(2 e^{x}-\pi/2)=\frac{-1}{\tan(2 e^{x})}$.

Use the formula for $\tan(2x)=\frac{2 \tan(x)}{1-(\tan(x))^2}$ and the result follows.

Answer 2

Hint: Use this fact that, under certain conditions, if $f'(x)=g'(x)$ then for a constant $c$: $f(x)-g(x)=c$.

You can see this for more.