Question

Simplify: \(\tan^{-1}\!\left(\dfrac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}}\right)\).

Hint:

Divide numerator and denominator by a^3 and match against the tan(3 theta) expansion formula.

Answer 1

Step 1: Key Formula:
Recall the triple-angle identity \(\tan3\phi=\dfrac{3\tan\phi-\tan^{3}\phi}{1-3\tan^{2}\phi}\).

Step 2: Matching the given expression:
Let \(\tan\phi=\dfrac{x}{a}\). Then \(\dfrac{3a^2x-x^3}{a^3-3ax^2}=\dfrac{a^3\big(3\tfrac{x}{a}-\tfrac{x^3}{a^3}\big)}{a^3\big(1-3\tfrac{x^2}{a^2}\big)}=\dfrac{3\tan\phi-\tan^{3}\phi}{1-3\tan^{2}\phi}=\tan3\phi\).

Final Answer:
So the expression equals \(\tan^{-1}(\tan3\phi)=3\phi=3\tan^{-1}\Big(\dfrac{x}{a}\Big)\).\[ \boxed{3\tan^{-1}\left(\dfrac{x}{a}\right)} \]