inverse function to find the image

Question

How can I prove that the image of

$f(x) = 2\tan(x) - 1/\cos(x)$

which is defined in $(-\pi/2, \pi/2)$ is $\mathbb{R}?$

I thought to find the inverse function of $f$ but I have no idea how to do it.

Answer 1

Let $t=\tan\frac x2 \in (-1,1)$. Then,

$$f(t) = \frac{4t}{1-t^2} - \frac{1+t^2}{1-t^2} = 1+\frac{4t-2}{1-t^2}$$

Note that $f(t)$ is continuous over $t\in(-1,1)$ and

$$\lim_{t\rightarrow -1} \frac{4t-2}{1-t^2} = -\infty,\>\>\>\>\> \lim_{t\rightarrow 1} \frac{4t-2}{1-t^2} = \infty,$$

Thus, $f(t)\in R$.