Write the function in the simplest form: tan | vedclass

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(A) Consider the expression $	an ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}ight)$.Let $x = a 	an 	heta$,which implies $\frac{x}{a} = 	an

Vedclass · Accepted Answer

$3 	an ^{-1} \frac{x}{a}$

Vedclass · Answer

(A) Consider the expression $	an ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}ight)$.Let $x = a 	an 	heta$,which implies $\frac{x}{a} = 	an 	heta$ or $	heta = 	an ^{-1}\left(\frac{x}{a}ight)$.Substituting $x = a 	an 	heta$ into the expression:$	an ^{-1}\left(\frac{3 a^{2} (a 	an 	heta) - (a 	an 	heta)^{3}}{a^{3} - 3 a (a 	an 	heta)^{2}}ight)$$= 	an ^{-1}\left(\frac{3 a^{3} 	an 	heta - a^{3} 	an ^{3} 	heta}{a^{3} - 3 a^{3} 	an ^{2} 	heta}ight)$$= 	an ^{-1}\left(\frac{a^{3}(3 	an 	heta - 	an ^{3} 	heta)}{a^{3}(1 - 3 	an ^{2} 	heta)}ight)$$= 	an ^{-1}\left(\frac{3 	an 	heta - 	an ^{3} 	heta}{1 - 3 	an ^{2} 	heta}ight)$Using the trigonometric identity $	an 3	heta = \frac{3 	an 	heta - 	an ^{3} 	heta}{1 - 3 	an ^{2} 	heta}$,we get:$= 	an ^{-1}(	an 3 	heta)$Since $\frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$,we have $\frac{-1}{\sqrt{3}} \leq 	an 	heta \leq \frac{1}{\sqrt{3}}$,which means $-\frac{\pi}{6} \leq 	heta \leq \frac{\pi}{6}$,so $-\frac{\pi}{2} \leq 3	heta \leq \frac{\pi}{2}$.Thus,$	an ^{-1}(	an 3 	heta) = 3 	heta = 3 	an ^{-1} \frac{x}{a}$.

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

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