Prove $\therefore\,2\,\tan^{\,-\,1}\,\frac{\sqrt{a^{\,2}- 2}}{a}= \tan^{\,-\,1}\,a\,\sqrt{a^{\,2}- 2}$

Prove $ $ \therefore\,2\,\tan^{\,-\,1}\,\frac{\sqrt{a^{\,2}- 2}}{a}= \tan^{\,-\,1}\,a\,\sqrt{a^{\,2}- 2} \tag{a p u}$ $ I find $ $ a\,\sqrt{a^{\,2}- 2}= \frac{\frac{\sqrt{a^{\,2}- 2}}{a}+ \frac{\sqrt{a^{\,2}- 2}}{a}}{1- \frac{\sqrt{a^{\,2}- 2}}{a}\,\frac{\sqrt{a^{\,2}- 2}}{a}}$ $ This may help! How I can use it to sol my problem! Thank you a real real lot!

Prove $\therefore\,2\,\tan^{\,-\,1}\,\frac{\sqrt{a^{\,2}- 2}}{a}= \tan^{\,-\,1}\,a\,\sqrt{a^{\,2}- 2}$

Prove $ $ \therefore\,2\,\tan^{\,-\,1}\,\frac{\sqrt{a^{\,2}- 2}}{a}= \tan^{\,-\,1}\,a\,\sqrt{a^{\,2}- 2} \tag{a p u}$ $ I find $ $ a\,\sqrt{a^{\,2}- 2}= \frac{\frac{\sqrt{a^{\,2}- 2}}{a}+ \frac{\sqrt{a^{\,2}- 2}}{a}}{1- \frac{\sqrt{a^{\,2}- 2}}{a}\,\frac{\sqrt{a^{\,2}- 2}}{a}}$ $ This may help! How I can use it to sol my problem! Thank you a real real lot!