sin ^{-1}left(frac{12}{13}right)+cos ^{-1}lef | vedclass

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(B) Let $	heta = \sin ^{-1}\left(\frac{12}{13}ight)$. Then $\sin 	heta = \frac{12}{13}$. Using the Pythagorean theorem, the base is $\sqrt{13^2 -

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$\pi$

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(B) Let $	heta = \sin ^{-1}\left(\frac{12}{13}ight)$. Then $\sin 	heta = \frac{12}{13}$. Using the Pythagorean theorem, the base is $\sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$. Thus, $	an 	heta = \frac{12}{5}$, so $	heta = 	an ^{-1}\left(\frac{12}{5}ight)$.Let $\phi = \cos ^{-1}\left(\frac{4}{5}ight)$. Then $\cos \phi = \frac{4}{5}$. The perpendicular is $\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$. Thus, $	an \phi = \frac{3}{4}$, so $\phi = 	an ^{-1}\left(\frac{3}{4}ight)$.Substituting these into the expression:$	an ^{-1}\left(\frac{12}{5}ight) + 	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{63}{16}ight)$Using the formula $	an ^{-1} A + 	an ^{-1} B = \pi + 	an ^{-1}\left(\frac{A+B}{1-AB}ight)$ for $AB > 1$:$A = \frac{12}{5}, B = \frac{3}{4} \Rightarrow AB = \frac{36}{20} = 1.8 > 1$.So, $	an ^{-1}\left(\frac{12}{5}ight) + 	an ^{-1}\left(\frac{3}{4}ight) = \pi + 	an ^{-1}\left(\frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} 	imes \frac{3}{4}}ight) = \pi + 	an ^{-1}\left(\frac{\frac{48+15}{20}}{1 - \frac{36}{20}}ight) = \pi + 	an ^{-1}\left(\frac{\frac{63}{20}}{-\frac{16}{20}}ight) = \pi + 	an ^{-1}\left(-\frac{63}{16}ight) = \pi - 	an ^{-1}\left(\frac{63}{16}ight)$.Adding the final term:$\pi - 	an ^{-1}\left(\frac{63}{16}ight) + 	an ^{-1}\left(\frac{63}{16}ight) = \pi$.

$\sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right)=$

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