Browse Questions

# Write the following function in the simplest form:$tan^{-1} \bigg( \frac{3a^2x-x^3}{a^3-3ax^2} \bigg), a > 0 ; \frac{-a}{\sqrt 3} \leq x \leq \frac{a}{\sqrt 3}$

Toolbox:
• $tan3\theta= \Large \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta}$
• $tan^{-1}tan\; \theta = \theta$
Given $tan^{-1} \bigg( \large \frac{3a^2x-x^3}{a^3-3ax^2} \bigg)$, $a > 0 ; \frac{-a}{\sqrt 3} \leq x \leq \frac{a}{\sqrt 3}$
Dividing the numerator and denominator by $a^3$, the given expression becomes $tan^{-1} \bigg( \Large \frac{3\frac{x}{a}-\frac{x^3}{a^3}}{1-3\frac{x^2}{a^2}}$$\bigg)$
Let $\large \frac{x}{a}$ $=tan\theta$ $\Rightarrow \theta = tan^{-1}\large \frac{x}{a}$
Substituting for $\frac{x}{a}$, the expression $tan^{-1} \bigg( \Large \frac{3\frac{x}{a}-\frac{x^3}{a^3}}{1-3\frac{x^2}{a^2}}$$\bigg)$$= tan^{-1} \bigg( \large \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta} \bigg)$
Since $tan3\theta= \Large \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta}$, this reduces to $tan^{-1}\: tan3\theta = 3\theta$
Since $\theta = tan^{-1}\large \frac{x}{a}$, $tan^{-1} \bigg( \large \frac{3a^2x-x^3}{a^3-3ax^2} \bigg)$ $=3\: tan^{-1}\frac{x}{a}$
edited Mar 14, 2013