Prove that $tan^{-1}frac{sqrt{1+x^2}+sqrt{1-x^2}}{sqrt{1+x^2}-sqrt{1-x^2}}=frac{pi}{4}+frac 12 cos^{-1}x^2$

A bit late answer but I thought worth mentioning it.

First note that we can substitute $y=x^2$ and consider $0<yleq 1$. Furthermore, the argument of $arctan$ can be simplified as follows:

$$frac{sqrt{1+y}+sqrt{1-y}}{sqrt{1+y}-sqrt{1-y}}=frac{1+sqrt{1-y^2}}{y}$$

Now, setting $y = cos t$ for $t in left[0,frac{pi}2right)$, to show is only

$$arctan frac{1+sin t}{cos t} = frac{pi}{4}+frac t2$$

At this point half-angle formulas come into mind:

$$frac{1+sin t}{cos t} = frac{(cos frac t2 + sin frac t2)^2}{cos^2 frac t2 - sin^2 frac t2} = frac{cos frac t2 + sin frac t2}{cos frac t2 - sin frac t2}$$ $$ = frac{1+tan frac t2}{1-tan frac t2} = tanleft(frac{pi}{4}+frac t2right)$$.

Done.

Your mistake

$sin y=astackrel{text{to}}{longrightarrow}sin^{-1}(sin y)=begin{cases}2npi+y&yintext{I, IV quadrant}\(2n-1)pi-y&yintext{II, III quadrant}end{cases}=sin^{-1}a$

Domain of $tan^{-1}frac{sqrt{1+x^2}+sqrt{1-x^2}}{sqrt{1+x^2}-sqrt{1-x^2}}$ is $x in (-1,1]$, nothing wrong with that.

But range of the above term(argument of $arctan$) is $(1,infty)$ so this means, when you assume it be $phi$, it is restricted to the interval $left[ frac{pi}{4},frac{pi}{2}right]$

This creates a problem in the last line, because $sin^{-1}(x^2)$ should be in its principal range and $frac{pi}{2}le 2phi le pi$

Edit: to find the domain of the argument, the easiest way, divide both sides by $sqrt{1+x^2}$ and then substitute $x^2=cos(2theta)$, $theta in left(0,frac{pi}{4}right)$, to get

$$frac{1+sqrt{frac{1-cos 2 theta}{1+cos 2 theta}}}{1-sqrt{frac{1-cos 2 theta}{1+cos 2 theta}}}$$

and then use the identity, $tan theta=sqrt{frac{1-cos 2 theta}{1+cos 2 theta}}$ $$frac{1+tan theta}{1-tan theta}=tan left(frac{pi}{4}+ thetaright)$$ Now, $theta in left(0,frac{pi}{4}right)$ so, $tan left(frac{pi}{4}+thetaright) in [1, infty)$

Which is ironically your question...

Because for $xneq0$ and $-1leq xleq1$ easy to see that: $$0<frac{pi}{4}+frac 12 cos^{-1}x^2<frac{pi}{2}$$ and we obtain: $$tanleft(frac{pi}{4}+frac 12 cos^{-1}x^2right)=frac{1+tanfrac{1}{2}arccos{x^2}}{1-tanfrac{1}{2}arccos{x^2}}=$$ $$=frac{cosfrac{1}{2}arccos{x^2}+sinfrac{1}{2}arccos{x^2}}{cosfrac{1}{2}arccos{x^2}-sinfrac{1}{2}arccos{x^2}}=frac{sqrt{frac{1+x^2}{2}}+sqrt{frac{1-x^2}{2}}}{sqrt{frac{1+x^2}{2}}-sqrt{frac{1-x^2}{2}}}=frac{sqrt{1+x^2}+sqrt{1-x^2}}{sqrt{1+x^2}-sqrt{1-x^2}}.$$

Your mistake in the last line.

Indeed, since $$frac{sqrt{1+x^2}+sqrt{1-x^2}}{sqrt{1+x^2}-sqrt{1-x^2}}>1$$ and from here $$frac{pi}{4}<phi<frac{pi}{2},$$ we obtain: $$2phi=pi-arcsin{x^2}=frac{pi}{2}+arccos{x^2}.$$