The value of tan left(sin ^{-1}left(frac{3}{5 | vedclass

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(B) We know that $\sin ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{3}{\sqrt{1^2 - (3/5)^2}}ight) = 	an ^{-1}\left(\frac{3/5}{4/5}ight)

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$\frac{17}{6}$

Vedclass · Answer

(B) We know that $\sin ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{3}{\sqrt{1^2 - (3/5)^2}}ight) = 	an ^{-1}\left(\frac{3/5}{4/5}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$.Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$,we get:$	an \left[	an ^{-1}\left(\frac{3}{4}ight)+	an ^{-1}\left(\frac{2}{3}ight)ight]$$= 	an \left[	an ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}ight)ight]$$= 	an \left[	an ^{-1}\left(\frac{\frac{9+8}{12}}{1-\frac{6}{12}}ight)ight]$$= 	an \left[	an ^{-1}\left(\frac{17/12}{6/12}ight)ight]$$= 	an \left[	an ^{-1}\left(\frac{17}{6}ight)ight]$$= \frac{17}{6}$

The value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

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