If $f(x)=\frac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$ let $f(x)=\dfrac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$
my try : $x:=\tan t$ then $f(\tan t)=\dfrac{\tan t}{1+\tan^2 t}=\sin t \ \ \  \text{or} \ \ -\sin t$
Now what ?
 A: Hint: use $f(f^{-1}(x)) = x$. $f^{-1}(x)$ can be solved explicitly in $x$. 
A: $f(x)=\dfrac{x}{\sqrt{x^2+1}}$
Let $y=f(x)$
Let's start by making x a function of y, which will give you inverse. 
$y=\dfrac{x}{\sqrt{x^2+1}}$
This simplifies to,
$y^2(x^2+1)=x^2$
From here, 
$x=\dfrac{y}{\sqrt{1-y^2}}$ 
Therefore,
$f^{-1}(x) = \dfrac{x}{\sqrt{1-x^2}}$
Hence, 
$f^{-1}(\sin x) = \dfrac{\sin x}{\sqrt{1-\sin^2 x}}=\tan x$
A: I would try to find the inverse of $f$ first - this can be done by switching the $x$ and $y$ variables, as follows:
Square to get rid of the square root:
\begin{align}
y = \frac{x}{\sqrt{x^2+1}} \Rightarrow y^2&=\frac{x^2}{x^2+1} \\
&= 1-\frac{1}{x^2+1}
\end{align}
then rearrange to get $x$ to be the subject variable:
\begin{align}
1-y^2=\frac{1}{x^2+1} &\Rightarrow x^2+1=\frac{1}{1-y^2}\\
&\Rightarrow x= \sqrt{\frac{1}{1-y^2}-1}
\end{align}
So $f^{-1}(x)$ can be defined as
$$
f^{-1}(x) = \sqrt{\frac{1}{1-x^2}-1}
$$
Now plugging in $\sin(x)$ yields
\begin{align}
\sqrt{\frac{1}{1-\sin^2(x)}-1} &= \sqrt{\frac{1}{\cos^2(x)}-1}\\
 &= \sqrt{\sec^2(x)-1}\\
&= \sqrt{\tan^2(x)}\\
&= \tan(x)
\end{align}
