Question

Write the given trigonometric expression in its simplest form.

${\tan }^{-1}\left(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right),a>0;\frac{-a}{\sqrt{3}}\le x\le \frac{a}{\sqrt{3}}$.

Answer 1

Consider the given trigonometric expression.

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

Substitute $x=a\tan \theta$.

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

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

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

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

Using the formula of $\tan 3\theta$, we have,

$\Rightarrow 3\theta$

Since,

$x=a\tan \theta$

$\frac{x}{a}=\tan \theta$

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

Therefore,

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

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

This is the simplest form of the given expression.