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(B) Let $\alpha = \sin^{-1}\left(\frac{3}{5}ight)$. Then $\sin \alpha = \frac{3}{5}$,which implies $	an \alpha = \frac{3}{4}$.Let $\beta = \cos^{-1

Vedclass · Accepted Answer

$\frac{-7}{24}$

Vedclass · Answer

(B) Let $\alpha = \sin^{-1}\left(\frac{3}{5}ight)$. Then $\sin \alpha = \frac{3}{5}$,which implies $	an \alpha = \frac{3}{4}$.Let $\beta = \cos^{-1}\left(\frac{2}{\sqrt{5}}ight)$. Then $\cos \beta = \frac{2}{\sqrt{5}}$,which implies $	an \beta = \frac{1}{2}$.We need to find $	an(\alpha - 2\beta)$.First,calculate $	an(2\beta) = \frac{2 	an \beta}{1 - 	an^2 \beta} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.Now,use the formula $	an(\alpha - 2\beta) = \frac{	an \alpha - 	an 2\beta}{1 + 	an \alpha 	an 2\beta}$.Substituting the values: $	an(\alpha - 2\beta) = \frac{3/4 - 4/3}{1 + (3/4)(4/3)} = \frac{(9-16)/12}{1 + 1} = \frac{-7/12}{2} = -\frac{7}{24}$.

Considering only the principal values of the inverse trigonometric functions,the value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$ is

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