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. 

Revised for the last part
In your specific case 
$$
\cos ^{ - 1} \left( {\frac{{x^2  - 1}}
{{x^2  + 1}}} \right) + \tan ^{ - 1} \left( {\frac{{2x}}
{{x^2  - 1}}} \right)
$$
you may note that:
$$
\left\{ \begin{gathered}
  x^2  - 1 = \operatorname{Re} \left( {\left( {x + i} \right)^2 } \right) \hfill \\
  2x = \operatorname{Im} \left( {\left( {x + i} \right)^2 } \right) \hfill \\
  \left( {x^2  + 1} \right) = \left| {\left( {x + i} \right)} \right|^2  = \left| {\left( {x + i} \right)^2 } \right| \hfill \\ 
\end{gathered}  \right.
$$
so that
$$
\begin{gathered}
 f(x)= \cos ^{ - 1} \left( {\frac{{x^2  - 1}}
{{x^2  + 1}}} \right) + \tan ^{ - 1} \left( {\frac{{2x}}
{{x^2  - 1}}} \right) =  \hfill \\
   = \cos ^{ - 1} \left( {\frac{{\operatorname{Re} \left( {\left( {x + i} \right)^2 } \right)}}
{{\left| {\left( {x + i} \right)^2 } \right|}}} \right) + \tan ^{ - 1} \left( {\frac{{\operatorname{Im} \left( {\left( {x + i} \right)^2 } \right)}}
{{\operatorname{Re} \left( {\left( {x + i} \right)^2 } \right)}}} \right) =  \hfill \\
   = \cos ^{ - 1} \left( {\cos \left( {2\arg \left( {x + i} \right)} \right)} \right) + \tan ^{ - 1} \left( {\tan \left( {2\arg \left( {x + i} \right)} \right)} \right) \hfill \\ 
\end{gathered} 
$$
Here we have to stop and do some case distinction, because
of the codomain limitation in the definition of $arccos$ and $arctan$, which if overlooked will cause the error signalled by Jean-Claude Arbaut.
In fact
$$
\cos ^{ - 1} \left( {\cos \alpha } \right) = \left| {\alpha _ *  } \right|\quad \quad \tan ^{ - 1} \left( {\tan \alpha } \right) = \alpha _{\, *  * } \quad \left| {\;\alpha _{\, *  * }  \ne \pi /2} \right.
$$
where the star indices indicate the necessary reductions mod $2\pi$ and $\pi$, that is
$$
\begin{gathered}
  \alpha _{\, * }  = \pi  - \bmod \left( {\pi  - \alpha ,\;2\,\,\pi } \right)\quad \left| \; \right. - \pi  < \alpha _{\, * }  \leqslant \pi  \hfill \\
  \alpha _{\, *  * }  = \pi /2 - \bmod \left( {\pi /2 - \alpha ,\;\,\pi } \right)\quad \left| \; \right. - \,\pi /2 < \alpha _{\, *  * }  \leqslant \pi /2 \hfill \\ 
\end{gathered} 
$$
So the development above continues as:
$$
\begin{gathered}
  f(x) = \cos ^{ - 1} \left( {\frac{{x^2  - 1}}
{{x^2  + 1}}} \right) + \tan ^{ - 1} \left( {\frac{{2x}}
{{x^2  - 1}}} \right)\quad \left| {\;x \ne 0,\; \pm 1} \right.\quad  =  \hfill \\
   = \cos ^{ - 1} \left( {\cos \left( {2\arg \left( {x + i} \right)} \right)} \right) + \tan ^{ - 1} \left( {\tan \left( {2\arg \left( {x + i} \right)} \right)} \right) =  \hfill \\
   = \left| {\left( {2\arg \left( {x + i} \right)} \right)_ *  } \right| + \left( {2\arg \left( {x + i} \right)} \right)_{\, *  * }  =  \hfill \\
   = \left| {\left( {2\tan ^{ - 1} \left( {1/x} \right)} \right)_ *  } \right| + \left( {2\tan ^{ - 1} \left( {1/x} \right)} \right)_{\, *  * }  \hfill \\ 
\end{gathered} 
$$
for finally arriving at
$$
f(x) = \left\{ {\begin{array}{*{20}c}
   0 & {x <  - 1}  \\
   \pi  & { - 1 < x \leqslant 0}  \\
   {4\tan ^{ - 1} \left( {1/x} \right) - \pi } & {0 < x < 1}  \\
   {4\tan ^{ - 1} \left( {1/x} \right)} & {1 < x}  \\
 \end{array} } \right.
$$
where the critical points have been filled according to the limit values
of $f(x)$.
