Q.Find the values of each of the following: $\tan^{-1} \left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)$, $a > 0$; $-\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$

Concept: Inverse Tangent Identity

We use the formula $\tan^{-1} \left( \frac{3t - t^3}{1 - 3t^2} \right) = 3 \tan^{-1} t$, valid when $t \in \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Step 1: Factor out $a^3$ from numerator and denominator

Step 2: Apply the identity …

We need to simplify $\tan^{-1} \left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)$ for $a > 0$ and $-\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$.

The key is recognizing the structure of the numerator and denominator. They strongly resemble the expansion of $\tan 3\theta$ in terms of $\tan \theta$.

Recall the triple-angle formula for tangent:

If we let $t = \tan\theta$, then $\tan 3\theta = \frac{3t - t^3}{1 - 3t^2}$.

Our expression has $\frac{3a^2x - x^3}{a^3 - 3ax^2}$. Factor $a^3$ from the denominator and $a^2$ from the numerator strategically:

1. Factor to match the standard form

   Numerator: $3a^2x - x^3 = a^3 \cdot \frac{3x}{a} - x^3$ — but better: factor $a^3$ from both numerator and denominator.

   Write numerator as $a^3 \left( \frac{3x}{a} - \frac{x^3}{a^3} \right) = a^3 \left( 3\left(\frac{x}{a}\right) - \left(\frac{x}{a}\right)^3 \right)$

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

2. Cancel the common factor $a^3$

   Since $a > 0$, $a^3 \neq 0$, so:

3. Recognize the triple-angle pattern Let $t = \frac{x}{a}$. Then the expression inside $\tan^{-1}$ becomes:

This is exactly $\tan 3\theta$ where $t = \tan\theta$.

4. Apply the inverse tangent identity For $t$ in a suitable range, $\tan^{-1}\left( \frac{3t - t^3}{1 - 3t^2} \right) = 3\tan^{-1} t$. …

Method: Recognising a tangent triple-angle pattern

Use this when a $\tan^{-1}$ argument is a ratio of cubics in $x$ and $a$.

Steps

Step 1: Factor out $a^3$ to expose $\frac{x}{a}$

Dividing numerator and denominator by $a^3$ turns the argument into $\frac{3t - t^3}{1 - 3t^2}$ where $t = \frac{x}{a}$.

Step 2: Identify the triple-angle formula

$\frac{3t - t^3}{1 - 3t^2} = \tan 3\phi$ when $t = \tan\phi$; here $\phi = \tan^{-1}\frac{x}{a}$. …

Common Mistakes

Mistake 1: Not dividing by $a^3$ to reveal the $\frac{x}{a}$ structure

Why it's wrong: without normalising by $a$, the ratio looks like an arbitrary cubic instead of the standard $\frac{3t - t^3}{1 - 3t^2}$. Correct approach: substitute $t = \frac{x}{a}$ to expose $\tan 3\phi$.

Mistake 2: Ignoring the domain and cancelling $\tan^{-1}(\tan 3\phi) = 3\phi$ blindly …