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$$\tan^{-1}\left(\dfrac{3a^2x-x^3}{a^3-3ax^2}\right),\;a>0;\; -\frac{a}{\sqrt{3}}\leqslant x \leqslant \frac{a}{\sqrt{3}}$$ Hi, Please help me to solve this problem. As I can solve simple inverse trigonometric functions. Please help hanks in advance.

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Just put $x = a \tan(\theta)$, then you can solve it.

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Recall the identity $$\tan(a+b)=\frac{\tan a +\tan b}{1-\tan a\tan b}.\tag{$1$}$$ Letting $a=t$ and $b=2t$ we find that $$\tan(3t)=\frac{\tan t+\tan 2t}{1-\tan t\tan 2t}.\tag{$2$}$$ But by using Identity $(1)$, we find that $$\tan 2t=\frac{2\tan t}{1-\tan^2 t}.$$ Substitute for $\tan 2t$ in $(2)$. After some simplification we find that $$\tan 3t=\frac{3\tan t -\tan^3 t}{1-3\tan^2 t}.$$

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