Find the value of tan left(sin ^{-1} frac{3}{ | vedclass

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Question

(A) Let $\sin ^{-1} \frac{3}{5} = x$. Then $\sin x = \frac{3}{5}$. Since $\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}

Vedclass · Accepted Answer

$\frac{17}{6}$

Vedclass · Answer

(A) Let $\sin ^{-1} \frac{3}{5} = x$. Then $\sin x = \frac{3}{5}$. Since $\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$,we have $	an x = \frac{\sin x}{\cos x} = \frac{3/5}{4/5} = \frac{3}{4}$. Thus,$\sin ^{-1} \frac{3}{5} = 	an ^{-1} \frac{3}{4}$. Also,$\cot ^{-1} \frac{3}{2} = 	an ^{-1} \frac{2}{3}$. Substituting these into the expression: $	an \left(	an ^{-1} \frac{3}{4} + 	an ^{-1} \frac{2}{3}ight)$. Using the formula $	an ^{-1} a + 	an ^{-1} b = 	an ^{-1} \left(\frac{a+b}{1-ab}ight)$: $= 	an \left(	an ^{-1} \left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \cdot \frac{2}{3}}ight)ight)$. $= 	an \left(	an ^{-1} \left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}ight)ight) = 	an \left(	an ^{-1} \left(\frac{17/12}{6/12}ight)ight)$. $= 	an \left(	an ^{-1} \frac{17}{6}ight) = \frac{17}{6}$.

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

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