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

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  • \( tan3\theta= \Large \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta}\)
  • $tan^{-1}tan\; \theta = \theta$
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 $\theta = tan^{-1}\large \frac{x}{a}$, $tan^{-1} \bigg( \large \frac{3a^2x-x^3}{a^3-3ax^2} \bigg)$ \( =3\: tan^{-1}\frac{x}{a} \)
answered Feb 22, 2013 by thanvigandhi_1
edited Mar 14, 2013 by balaji.thirumalai

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