sin left( 4 tan^{-1} frac{1}{3} right) = | vedclass

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(B) Let $	heta = 	an^{-1} \frac{1}{3}$,then $	an 	heta = \frac{1}{3}$.We need to find $\sin(4	heta)$.First,calculate $	an(2	heta) = \frac{2 	a

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$\frac{24}{25}$

Vedclass · Answer

(B) Let $	heta = 	an^{-1} \frac{1}{3}$,then $	an 	heta = \frac{1}{3}$.We need to find $\sin(4	heta)$.First,calculate $	an(2	heta) = \frac{2 	an 	heta}{1 - 	an^2 	heta} = \frac{2(1/3)}{1 - (1/9)} = \frac{2/3}{8/9} = \frac{2}{3} 	imes \frac{9}{8} = \frac{3}{4}$.Now,calculate $\sin(4	heta) = \sin(2(2	heta)) = \frac{2 	an(2	heta)}{1 + 	an^2(2	heta)}$.Substituting $	an(2	heta) = \frac{3}{4}$:$\sin(4	heta) = \frac{2(3/4)}{1 + (3/4)^2} = \frac{3/2}{1 + 9/16} = \frac{3/2}{25/16} = \frac{3}{2} 	imes \frac{16}{25} = \frac{24}{25}$.Thus,the correct option is $B$.

$\sin \left( 4 \tan^{-1} \frac{1}{3} \right) = $

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