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 ?

$f(x)=\dfrac{x}{\sqrt{x^2+1}}$
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, 
$f^{-1}(\sin x) = \dfrac{\sin x}{\sqrt{1-\sin^2 x}}=\tan x$
