Prove that $2\tan^{-1}\frac{\sqrt{a^{2}- 2}}{a}= \tan^{-\,1}a\sqrt{a^{2}- 2} $ Prove
$$2\tan^{-1}\dfrac{\sqrt{a^{2}- 2}}{a}= \tan^{-\,1}a\sqrt{a^{2}- 2} \tag{a p u}$$
I find
$$a\sqrt{a^{2}- 2}= \dfrac{\dfrac{\sqrt{a^{2}- 2}}{a}+ \dfrac{\sqrt{a^{2}- 2}}{a}}{1- \dfrac{\sqrt{a^{2}- 2}}{a}\dfrac{\sqrt{a^{2}- 2}}{a}}$$
This may help! How I can use it to solve my problem? Thanks for all the interests!
 A: Using 
$$\tan^{-1}(u) - \tan^{-1}(v) = \tan^{-1} \frac{u - v}{1 + uv}$$
where $u=\frac{\sqrt{a^2-2}}{a}$ and $v = a\sqrt{a^2-  2}$
you get 
$$\tan^{-1}\frac{\sqrt{a^2-2}}{a} - \tan^{-1} a\sqrt{a^2-  2} = \tan^{-1} \frac{\frac{\sqrt{a^2-2}}{a} - a\sqrt{a^2-2}}{1+a^2-2} = -\tan^{-1}\frac{\sqrt{a^2-2}}{a}  $$
A: First, note that, whenever $\frac{\sqrt{a^2 - 2}}{a}$ is well-defined, we have
$$-1 < \frac{\sqrt{a^2 - 2}}{a} < 1,$$
and since $\tan^{-1}$ is an increasing function,
$$-\frac{\pi}{4} = \tan^{-1}(-1) < \tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right) < \tan^{-1}(1) = \frac{\pi}{4}.$$
Therefore,
$$-\frac{\pi}{2} < 2\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right) < \frac{\pi}{2}.$$
For angles $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, we have $\tan^{-1} (\tan \theta) = \theta$. Also note that, for any $\alpha \in \Bbb{R}$, we have $\tan(\tan^{-1}(\alpha)) = \alpha$. Thus, by the double angle formula for $\tan$,
\begin{align*}
2\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right) &= \tan^{-1}\left(\tan\left(2\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right)\right)\right) \\
&= \tan^{-1}\left(\frac{2\tan \left(\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right)\right)}{1 - \tan \left(\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right)\right)^2}\right) \\
&= \tan^{-1}\left(\frac{2\frac{\sqrt{a^2 - 2}}{a}}{1 - \left(\frac{\sqrt{a^2 - 2}}{a}\right)^2}\right) \\
&= \tan^{-1}\left(a\sqrt{a^2 - 2}\right).
\end{align*}
A: HINT
Recall that

$$\tan^{-1}(x)+\tan^{-1}(y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$

In your case $x=y$; and now take a closer look at your simplification.
