tan left(cos ^{-1}left(frac{4}{5}right)+tan ^ | vedclass

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Question

(A) Let $	heta_1 = \cos^{-1}\left(\frac{4}{5}ight)$. Then $\cos 	heta_1 = \frac{4}{5}$.Using the identity $	an 	heta_1 = \frac{\sqrt{1-\cos^2

Vedclass · Accepted Answer

$\frac{17}{6}$

Vedclass · Answer

(A) Let $	heta_1 = \cos^{-1}\left(\frac{4}{5}ight)$. Then $\cos 	heta_1 = \frac{4}{5}$.Using the identity $	an 	heta_1 = \frac{\sqrt{1-\cos^2 	heta_1}}{\cos 	heta_1} = \frac{\sqrt{1-(16/25)}}{4/5} = \frac{3/5}{4/5} = \frac{3}{4}$.So,$\cos^{-1}\left(\frac{4}{5}ight) = 	an^{-1}\left(\frac{3}{4}ight)$.Now,the expression becomes $	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)$:$= 	an \left(	an^{-1} \left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} 	imes \frac{2}{3}}ight)ight)$.$= 	an \left(	an^{-1} \left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}ight)ight) = 	an \left(	an^{-1} \left(\frac{17/12}{6/12}ight)ight) = 	an \left(	an^{-1} \left(\frac{17}{6}ight)ight) = \frac{17}{6}$.

$\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) = $

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