Question
Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),\;a > 0,\left( { - \frac{a}{{\sqrt 3 }} < x < \frac{a}{{\sqrt 3 }}} \right)$

Answer

${\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{3\left( {\frac{x}{a}} \right) - {{\left( {\frac{x}{a}} \right)}^3}}}{{1 - 3{{\left( {\frac{x}{a}} \right)}^2}}}} \right) [$Dividing numerator and denominator by $a^3]$
Putting $\frac{x}{a} = \tan \theta$ so that $\theta = {\tan ^{ - 1}}\frac{x}{a}$
$ = {\tan ^{ - 1}}\left( {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)$
$ = {\tan ^{ - 1}}\tan 3\theta$
$ = 3\theta = 3{\tan ^{ - 1}}\frac{x}{a}$

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