What is the value of sin ^{-1} frac{12}{13}+c | vedclass

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(A) Let $x = \sin ^{-1} \frac{12}{13} + \cos ^{-1} \frac{4}{5} + 	an ^{-1} \frac{63}{16}$.Convert all terms to $	an ^{-1}$ form:$\sin ^{-1} \frac{12

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$\pi$

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(A) Let $x = \sin ^{-1} \frac{12}{13} + \cos ^{-1} \frac{4}{5} + 	an ^{-1} \frac{63}{16}$.Convert all terms to $	an ^{-1}$ form:$\sin ^{-1} \frac{12}{13} = 	an ^{-1} \frac{12}{5}$ (since $\sin 	heta = \frac{12}{13} \implies 	an 	heta = \frac{12}{5}$)$\cos ^{-1} \frac{4}{5} = 	an ^{-1} \frac{3}{4}$ (since $\cos 	heta = \frac{4}{5} \implies 	an 	heta = \frac{3}{4}$)Now,the expression becomes $	an ^{-1} \frac{12}{5} + 	an ^{-1} \frac{3}{4} + 	an ^{-1} \frac{63}{16}$.Using $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \frac{x+y}{1-xy}$:$	an ^{-1} \frac{12}{5} + 	an ^{-1} \frac{3}{4} = 	an ^{-1} \left( \frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \cdot \frac{3}{4}} ight) = 	an ^{-1} \left( \frac{\frac{48+15}{20}}{1 - \frac{36}{20}} ight) = 	an ^{-1} \left( \frac{63/20}{-16/20} ight) = 	an ^{-1} \left( -\frac{63}{16} ight) = -	an ^{-1} \frac{63}{16}$.Substituting this back: $-	an ^{-1} \frac{63}{16} + 	an ^{-1} \frac{63}{16} = 0$.Wait,let us re-evaluate the sum: $	an ^{-1} \frac{12}{5} + 	an ^{-1} \frac{3}{4} = \pi + 	an ^{-1} \left( -\frac{63}{16} ight) = \pi - 	an ^{-1} \frac{63}{16}$ (since the product $xy > 1$).Thus,$(\pi - 	an ^{-1} \frac{63}{16}) + 	an ^{-1} \frac{63}{16} = \pi$.

What is the value of $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}$?

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