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

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(B) Let $	an ^{-1} \frac{3}{4} = 	heta$. Then $	an 	heta = \frac{3}{4}$.Using the identity $\sin 	heta = \frac{	an 	heta}{\sqrt{1 + 	an^2 	he

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(B) Let $	an ^{-1} \frac{3}{4} = 	heta$. Then $	an 	heta = \frac{3}{4}$.Using the identity $\sin 	heta = \frac{	an 	heta}{\sqrt{1 + 	an^2 	heta}}$,we get $\sin 	heta = \frac{3/4}{\sqrt{1 + (3/4)^2}} = \frac{3/4}{\sqrt{25/16}} = \frac{3/4}{5/4} = \frac{3}{5}$.Thus,$\left[\sin \left(	an ^{-1} \frac{3}{4}ight)ight]^2 = \left(\frac{3}{5}ight)^2 = \frac{9}{25}$.Let $	an ^{-1} \frac{4}{3} = \phi$. Then $	an \phi = \frac{4}{3}$.Using the identity $\sin \phi = \frac{	an \phi}{\sqrt{1 + 	an^2 \phi}}$,we get $\sin \phi = \frac{4/3}{\sqrt{1 + (4/3)^2}} = \frac{4/3}{\sqrt{25/9}} = \frac{4/3}{5/3} = \frac{4}{5}$.Thus,$\left[\sin \left(	an ^{-1} \frac{4}{3}ight)ight]^2 = \left(\frac{4}{5}ight)^2 = \frac{16}{25}$.Adding the two values: $\frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$.

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

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