<s>[INST] 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!
 [/INST] 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 
$$\tan^{-1}\frac{\sqrt{a^2-2}}{a} - \tan^{-1} a\sqrt{a^2-  2} = \tan^{-1} \frac{\frac{\sqrt{a^2-2}}{a} - a\sqrt{a^2-2}}{1+a^2-2} = -\tan^{-1}\frac{\sqrt{a^2-2}}{a}  $$
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