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Mathematics Question on Inverse Trigonometric Functions

Write the function in the simplest form: tan⁡−1(3a2x−x3a3−3ax2),a>0;−a3≤x≤a3\tan ^{-1}(\frac{3a^2x-x^3}{a^3-3ax^2}),a>0;\frac {-a}{\sqrt3} \leq x\leq\frac{ a}{\sqrt3}

Answer

tan⁡−1(3a2x−x3a3−3ax2)\tan ^{-1}(\frac{3a^2x-x^3}{a^3-3ax^2})

put x= atan⁡θ⇒xa=tan⁡θ⇒θ=tan⁡−1xa.a \tan\theta\Rightarrow\frac{x}{a}=\tan\theta\Rightarrow\theta=\tan^{-1}\frac{x}{a}.

= tan⁡−1(3a2x−x3a3−3ax2)=tan⁡−1(3a2.atan⁡θ−a.3tan⁡θa3−3a3tan⁡2θ)\tan^{-1}\bigg(\frac{3a^2x-x^3}{a^3-3ax^2}\bigg)=\tan^{-1}\bigg(\frac{3a^2.a\tan\theta-a.3\tan\theta}{a^3-3a^3\tan^2\theta}\bigg)

=tan⁡−1(3a33tan⁡θ−a.3tan⁡θa3−3a3tan⁡2θ)\tan^{-1}\bigg(\frac{3a^33\tan\theta-a.3\tan\theta}{a^3-3a^3\tan^2\theta}\bigg)

=tan⁡−1(3tan⁡θ−tan⁡3θ1−3tan⁡2θ)=tan⁡−1(tan⁡3θ)\tan^{-1}\bigg(\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\bigg)=\tan^{-1}(\tan3\theta)

=3θ

= 3tan⁡−1xa3\tan^{-1}\frac{x}{a}