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

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(D) Let $\alpha = \sin^{-1} \left( \frac{3}{5} ight)$. Then $\sin \alpha = \frac{3}{5}$,so $	an \alpha = \frac{3}{\sqrt{5^2 - 3^2}} = \frac{3}{4}$.

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$\frac{17}{6}$

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(D) Let $\alpha = \sin^{-1} \left( \frac{3}{5} ight)$. Then $\sin \alpha = \frac{3}{5}$,so $	an \alpha = \frac{3}{\sqrt{5^2 - 3^2}} = \frac{3}{4}$. Thus,$\alpha = 	an^{-1} \left( \frac{3}{4} ight)$.Let $\beta = \cos^{-1} \left( \frac{3}{\sqrt{13}} ight)$. Then $\cos \beta = \frac{3}{\sqrt{13}}$,so $	an \beta = \frac{\sqrt{(\sqrt{13})^2 - 3^2}}{3} = \frac{\sqrt{13-9}}{3} = \frac{2}{3}$. Thus,$\beta = 	an^{-1} \left( \frac{2}{3} ight)$.Now,the expression becomes $	an(\alpha + \beta) = 	an \left( 	an^{-1} \frac{3}{4} + 	an^{-1} \frac{2}{3} ight)$.Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$	an(\alpha + \beta) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left( \frac{3}{4} 	imes \frac{2}{3} ight)} = \frac{\frac{9+8}{12}}{1 - \frac{6}{12}} = \frac{\frac{17}{12}}{\frac{6}{12}} = \frac{17}{6}$.

The value of $\tan \left[ \sin^{-1} \left( \frac{3}{5} \right) + \cos^{-1} \left( \frac{3}{\sqrt{13}} \right) \right]$ is

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