tan left[ cos^{-1} frac{4}{5} + tan^{-1} frac | vedclass

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(B) Let $	heta = \cos^{-1} \frac{4}{5}$. Then $\cos 	heta = \frac{4}{5}$.Since $	an 	heta = \frac{\sqrt{1 - \cos^2 	heta}}{\cos 	heta} = \frac{\

Vedclass · Accepted Answer

$17/6$

Vedclass · Answer

(B) Let $	heta = \cos^{-1} \frac{4}{5}$. Then $\cos 	heta = \frac{4}{5}$.Since $	an 	heta = \frac{\sqrt{1 - \cos^2 	heta}}{\cos 	heta} = \frac{\sqrt{1 - (16/25)}}{4/5} = \frac{3/5}{4/5} = \frac{3}{4}$,we have $	heta = 	an^{-1} \frac{3}{4}$.Substituting this into the expression,we get $	an \left[ 	an^{-1} \frac{3}{4} + 	an^{-1} \frac{2}{3} ight]$.Using the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x+y}{1-xy} ight)$,we have:$	an \left[ 	an^{-1} \left( \frac{3/4 + 2/3}{1 - (3/4)(2/3)} ight) ight]$$= 	an \left[ 	an^{-1} \left( \frac{9/12 + 8/12}{1 - 6/12} ight) ight]$$= 	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}$.

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

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