Trig identity problem using inverse functions

Question

I'm looking to solve the following exercise.

Show that $\tan^{-1}(\sqrt{e^{2x}-1})=\sec^{-1}(e^x)$

essentially I must simplify the left side of the equation but I'm not sure where to start. Any tips?

Answer 1

Let $e^x =\sec\alpha$

So,$$\tan^{-1}\left(\sqrt{e^{2x} - 1}\right)= \tan^{-1}\left(\sqrt{(e^x)^2 - 1}\right) = \tan^{-1}\left(\sqrt{\sec^2\alpha - 1}\right) = \tan^{-1}(\tan\alpha) = \alpha$$

From our first consideration, $\alpha =\sec^{-1}(e^x)$

Thus, $$\tan^{-1}\left(\sqrt{e^{2x} - 1}\right) = \sec^{-1}(e^x)$$

Answer 2

We have to prove that:

$$\arctan(\sqrt{e^{2x}-1})-\text{arcsec}(e^x)=0$$

If $f(x)=\arctan(\sqrt{e^{2x}-1})-\text{arcsec}(e^x)$, then:

$$f'(x)=\frac{1}{\sqrt{e^{2x}-1}}-\frac{e^{-x}}{\sqrt{1-e^{-2x}}} $$

Do some calculus and you get:

$$f'(x)=0$$

This means that the function is constant in particular if you try $x=0$ you get $f(0)=0$ , this means that $f(x)=0$

:)