{left[ {sin left( {{{tan }^{ - 1}}frac{3}{4}} | vedclass

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Question

(C) Let $	heta = 	an^{-1}\left(\frac{3}{4}ight)$.Then,$	an 	heta = \frac{3}{4}$.In a right-angled triangle,if the opposite side is $3$ and the a

Vedclass · Accepted Answer

$\frac{9}{25}$

Vedclass · Answer

(C) Let $	heta = 	an^{-1}\left(\frac{3}{4}ight)$.Then,$	an 	heta = \frac{3}{4}$.In a right-angled triangle,if the opposite side is $3$ and the adjacent side is $4$,then the hypotenuse is $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.Therefore,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{3}{5}$.Substituting this back into the expression: ${\left[ \sin \left( 	an^{-1} \frac{3}{4} ight) ight]^2} = (\sin 	heta)^2 = \left( \frac{3}{5} ight)^2 = \frac{9}{25}$.

${\left[ {\sin \left( {{{\tan }^{ - 1}}\frac{3}{4}} \right)} \right]^2} = $

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