Function : Find the range of $f(x) = \sin^{-1}x +\tan^{-1}x +\cos^{-1}x$ Problem : 
Find the range of $f(x) = \sin^{-1}x +\tan^{-1}x +\cos^{-1}x$
Solution : Since, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
Since range of $\tan^{-1}x$  is $ (\frac{-\pi}{2}, \frac{\pi}{2})$
$\therefore, \frac{-\pi}{2} \leq \tan^{-1}x \leq \frac{\pi}{2}$
= $ \frac{-\pi}{2} + \frac{\pi}{2} \leq \tan^{-1}x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$.
= $0 \leq \tan^{-1}x+ \frac{\pi}{2} \leq \pi $
Is it correct.. please suggest thanks.... 

This is not correct because you need to restrict $\tan^{-1}$ to the same domain as the other two functions: $[-1,1]$. The extreme vales that  $\tan^{-1}(x)$ takes on $[-1,1]$ are $\pm\frac{\pi}{4}$. So the range of your $f$ is $\left[\frac{\pi}{2}-\frac{\pi}{4},\frac{\pi}{2}+\frac{\pi}{4}\right]=\left[\frac{\pi}{4},\frac{3\pi}{4}\right]$ (and its domain is $[-1,1]$).

