Find the Range and Inverse of the function $f(x)=\sec x\tan x+\sec^2 x$

Question

Find the Range and Inverse of the function $f(x)=\sec x\tan x+\sec^2 x$

My try:

We have $$y=\sec x\left(\sec x+\tan x\right)$$

$$2y=\left(\sec x+\tan x+\sec x-\tan x\right)\left(\sec x+\tan x\right)$$

$$2y=\left(\sec x+\tan x\right)^2+1$$

So we get:

$$\sec x+\tan x=\sqrt{2y-1}$$ Also

$$\sec x-\tan x-\frac{1}{\sqrt{2y-1}}$$

Adding above two results we get:

$$x=\sec^{-1}\left(\frac{\sqrt{2y-1}+\frac{1}{\sqrt{2y-1}}}{2}\right)$$

Hence $$f^{-1}(x)=\sec^{-1}\left(\frac{\sqrt{2x-1}+\frac{1}{\sqrt{2x-1}}}{2}\right)$$

How to find Range?

Answer 1

Hint:

For $\cos x\ne0\iff\sin x\ne\pm1$ $$f(x)=\dfrac1{1-\sin x}$$

Here $0<1-\sin x<2\iff f(x)>\dfrac12$

If $y=f(x)=\cdots,$ $$f^{-1}(y)=x=\arcsin\left(1-\dfrac1y\right)$$