How to show that the function $f(x) = \sin x + \tan x$ on $(-\pi/2, \pi/2)$ has an inverse Given that $f(x)=\sin x+\tan x$ on $(-\pi/2,\pi/2)$. Show that it has an inverse function.  Find the derivative of the inverse of $f(x)$ at $x=0$.
What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?
 A: $\star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-\frac{\pi}{2},\frac{\pi}{2}[$.
$\star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}\neq g^{-1}+h^{-1}$$
But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$$
A: In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective. 
But, $\sin x$ and $\tan x$ are strictly increasing over $\left( -\frac{\pi}{2}, \frac{\pi}{2}\right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $\implies$ $f$ can be inversed.
Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand). 
But, consider the following standard Theorem proved in pre-calculus courses :

Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
  $$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$$
Proof :
Let $y_0\in \mathbb{R}$. Then $\exists x_0\in\mathbb{R} : f(x_0)=y_0$. Then $x\to x_0 \implies y=f(x)\to f(x_0)=y_0$.
  By definition, it is :
  $$(f^{-1})'(y_0)=\lim_{y\rightarrow y_0}\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=\lim_{y\rightarrow y_0}\frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\\=\lim_{x\rightarrow x_0}\frac{x-x_0}{f(x)-f(x_0)}=\frac{1}{f'(x_0)}=\frac{1}{f'(f^{-1}(y_0))}$$

Can you now use that theorem to find the derivative of the inverse at $x=0$ ?
