Simplifying inverse trigonometric expressions such as $\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{2x}{x^2-1}\right)$ Okay so I'm just looking for a short cut method or a method that is not so long for simplifying expressions like this
$$\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{2x}{x^2-1}\right)$$
Here is how I do it: I take the first term of the equation and convert it to $\tan^{-1}$ form, and then I apply the formula of $\tan^{-1}a +   \tan^{-1}b$. But this method is really time consuming; plus, it results in mistakes most of the times. Any alternatives or how you would do it, let me know, I'll be grateful. Thanking in anticipation. 
Edit. Do I put $x= \tan y$ and solve? That should make it easy. 

$$
\frac{1-\tan^2(u)}{1+\tan^2(u)}=\cos(2u)\implies
\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)
=\left\{\begin{array}{}
\pi+2\tan^{-1}(x)&\text{if }x\lt0\\
\pi-2\tan^{-1}(x)&\text{if }x\gt0
\end{array}\right.
$$
$$
\frac{2\tan(u)}{1-\tan^2(u)}=\tan(2u)\implies
\tan^{-1}\left(\frac{2x}{x^2-1}\right)
=\left\{\begin{array}{}
-\pi-2\tan^{-1}(x)&\text{if }x\lt-1\\
-2\tan^{-1}(x)&\text{if }-1\lt x\lt1\\
\pi-2\tan^{-1}(x)&\text{if }x\gt1
\end{array}\right.
$$
Therefore,
$$
\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{2x}{x^2-1}\right)
=\left\{\begin{array}{}
0&\text{if }x\lt-1\\
\pi&\text{if }-1\lt x\le0\\
\pi-4\tan^{-1}(x)&\text{if }0\lt x\lt1\\
2\pi-4\tan^{-1}(x)&\text{if }x\gt1
\end{array}\right.
$$
