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Question

(D) We know the trigonometric identity for $	an 3	heta$ is given by:$	an 3	heta = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$Let $\

Vedclass · Accepted Answer

$	an^{-1} \frac{3a - a^3}{1 - 3a^2}$

Vedclass · Answer

(D) We know the trigonometric identity for $	an 3	heta$ is given by:$	an 3	heta = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$Let $	an 	heta = a$,then $	heta = 	an^{-1} a$.Substituting this into the identity,we get:$	an 3(	an^{-1} a) = \frac{3a - a^3}{1 - 3a^2}$Taking $	an^{-1}$ on both sides,we obtain:$3 	an^{-1} a = 	an^{-1} \left( \frac{3a - a^3}{1 - 3a^2} ight)$Thus,the correct option is $D$.

$3 \tan^{-1} a$ is equal to

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