tan left( 2{{cos }^{ - 1}}frac{3}{5} right) = | vedclass

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(C) Let $	heta = \cos^{-1}\left(\frac{3}{5}ight)$. Then $\cos 	heta = \frac{3}{5}$.Since $\cos 	heta = \frac{3}{5}$,we have $\sin 	heta = \sqrt{

Vedclass · Accepted Answer

$-\frac{24}{7}$

Vedclass · Answer

(C) Let $	heta = \cos^{-1}\left(\frac{3}{5}ight)$. Then $\cos 	heta = \frac{3}{5}$.Since $\cos 	heta = \frac{3}{5}$,we have $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \left(\frac{3}{5}ight)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{4/5}{3/5} = \frac{4}{3}$.We need to find $	an(2	heta)$.Using the formula $	an(2	heta) = \frac{2	an 	heta}{1 - 	an^2 	heta}$,we get:$	an(2	heta) = \frac{2(4/3)}{1 - (4/3)^2} = \frac{8/3}{1 - 16/9} = \frac{8/3}{-7/9} = \frac{8}{3} 	imes \left(-\frac{9}{7}ight) = -\frac{24}{7}$.

$\tan \left( 2{{\cos }^{ - 1}}\frac{3}{5} \right) = $

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