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(B) Let $	heta_1 = \sin^{-1}(\frac{3}{5})$. Then $\sin 	heta_1 = \frac{3}{5}$,which implies $	an 	heta_1 = \frac{3}{4}$,so $	heta_1 = 	an^{-1}(\

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$\frac{-7}{24}$

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(B) Let $	heta_1 = \sin^{-1}(\frac{3}{5})$. Then $\sin 	heta_1 = \frac{3}{5}$,which implies $	an 	heta_1 = \frac{3}{4}$,so $	heta_1 = 	an^{-1}(\frac{3}{4})$.Let $	heta_2 = \cos^{-1}(\frac{2}{\sqrt{5}})$. Then $\cos 	heta_2 = \frac{2}{\sqrt{5}}$,which implies $	an 	heta_2 = \frac{1}{2}$,so $	heta_2 = 	an^{-1}(\frac{1}{2})$.The expression becomes $	an(	heta_1 - 2	heta_2) = 	an(	an^{-1}(\frac{3}{4}) - 2	an^{-1}(\frac{1}{2}))$.Using $2	an^{-1}(x) = 	an^{-1}(\frac{2x}{1-x^2})$,we get $2	an^{-1}(\frac{1}{2}) = 	an^{-1}(\frac{2 	imes \frac{1}{2}}{1 - \frac{1}{4}}) = 	an^{-1}(\frac{1}{3/4}) = 	an^{-1}(\frac{4}{3})$.Now,$	an(	an^{-1}(\frac{3}{4}) - 	an^{-1}(\frac{4}{3})) = 	an(	an^{-1}(\frac{3/4 - 4/3}{1 + (3/4)(4/3)})) = \frac{3/4 - 4/3}{1 + 1} = \frac{(9-16)/12}{2} = \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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