The value of tan left[cos ^{-1}left(frac{4}{5 | vedclass

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(A) We need to find the value of $	an \left[\cos ^{-1}\left(\frac{4}{5}ight)+	an ^{-1}\left(\frac{2}{3}ight)ight]$.Let $\cos ^{-1}\left(\frac{

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$\frac{17}{6}$

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(A) We need to find the value of $	an \left[\cos ^{-1}\left(\frac{4}{5}ight)+	an ^{-1}\left(\frac{2}{3}ight)ight]$.Let $\cos ^{-1}\left(\frac{4}{5}ight) = 	heta$. Then $\cos 	heta = \frac{4}{5}$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \left(\frac{4}{5}ight)^2 = 1 - \frac{16}{25} = \frac{9}{25}$,we have $\sin 	heta = \frac{3}{5}$.Therefore,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{3/5}{4/5} = \frac{3}{4}$.So,$	heta = 	an ^{-1}\left(\frac{3}{4}ight)$.Substituting this into the expression,we get $	an \left[	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{2}{3}ight)ight]$.Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left(\frac{x+y}{1-xy}ight)$ for $xy $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{2}{3}ight) = 	an ^{-1} \left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4}ight)\left(\frac{2}{3}ight)}ight) = 	an ^{-1} \left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}ight) = 	an ^{-1} \left(\frac{17/12}{6/12}ight) = 	an ^{-1} \left(\frac{17}{6}ight)$.Finally,$	an \left[	an ^{-1} \left(\frac{17}{6}ight)ight] = \frac{17}{6}$.

The value of $\tan \left[\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]$ is

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