Inverse Trigonometric Functions — Class 12 Maths Solution

Putting $x = a\;\;\tan \theta ,$ we get
${\tan ^{ - 1}}\left[ {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right] = {\tan ^{ - 1}}\left[ {\frac{{3{a^3}\tan \theta - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right]$

$= {\tan ^{ - 1}}\left( {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right) = {\tan ^{ - 1}}(\tan 3\theta ) = 3\theta = 3{\tan ^{ - 1}}\left( {\frac{x}{a}} \right)$