I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$ I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$
The easiest way was to just look at the graph and I found out that the region is $x \in ({1\over \sqrt{2}} , 1]$
But I tried to prove the statement algebraically also but couldn't get it correctly.
For $$\sin ^{-1}x > \cos ^{-1}x$$
$$\Rightarrow \sin ^{-1}x -\cos ^{-1}x>0$$
$$\Rightarrow \sin ^{-1}x - \sin ^{-1} \sqrt{1-x^2}>0 \tag1$$ 
Using the identity $$\sin ^{-1}x -\sin ^{-1}y=\sin ^{-1} \left (x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)$$ Equation $(1)$ can be written as
$$\Rightarrow \sin ^{-1}(2x^2-1)>0$$
For this to be true $0<2x^2-1<1$ and also for the equation to be valid $-1<2x^2-1<1$
Hence we can take 
$0<2x^2-1<1$ as the intersection of both the conditions
$$\Rightarrow 0\le  2x^2-1\le 1$$
$$\Rightarrow 1\le 2x^2\le 2$$
$$\Rightarrow 1/2\le x^2\le 1$$
The solution set of the inequality would be $x \in ({1\over \sqrt{2}} , 1] \cup [-1,{-1\over \sqrt{2}}),$
Can anybody tell me why I am getting the wrong answer.

About your method you have to note that, unfortunately, $\arccos x=\arcsin\sqrt{1-x^2}$ is a false identity in general; it's true only for $0\le x\le 1$.
However, for $-1\le x<0$ we have $-\pi/2\le\arcsin x<0$ and $\pi/2<\arccos x\le\pi$, so we can exclude this interval, where the inequality certainly doesn't hold.
So we have
$$
\begin{cases}
\arcsin x>\arcsin\sqrt{1-x^2} \\[4px]
0\le x\le 1
\end{cases}
$$
that becomes, since the arcsine is increasing
$$
\begin{cases}
x>\sqrt{1-x^2} \\[4px]
0\le x\le 1
\end{cases}
$$
Due to the second limitation, we can square the first inequality
$$
\begin{cases}
x^2>1-x^2 \\[4px]
0\le x\le 1
\end{cases}
$$
and easily get $1/\sqrt{2} < x \le 1$.
On the other hand, if $\alpha=\arccos x$, we have $0\le\alpha\le\pi$, so $-\pi/2\le\pi/2-\alpha\le\pi/2$ and from $x=\cos\alpha$ we get $x=\sin(\pi/2-\alpha)$, so
$$
\frac{\pi}{2}-\alpha=\arcsin x
$$
and therefore
$$
\arccos x=\frac{\pi}{2}-\arcsin x
$$
which makes your inequality very easy to solve.
