cos ^{ - 1}frac{4}{5} + tan ^{ - 1}frac{3}{5} | vedclass

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Question

(A) Let $x = \cos ^{ - 1}\frac{4}{5}$. Then $\cos x = \frac{4}{5}$.Since $\cos x = \frac{	ext{adj}}{	ext{hyp}} = \frac{4}{5}$,the opposite side is $

Vedclass · Accepted Answer

$	an ^{ - 1}\frac{27}{11}$

Vedclass · Answer

(A) Let $x = \cos ^{ - 1}\frac{4}{5}$. Then $\cos x = \frac{4}{5}$.Since $\cos x = \frac{	ext{adj}}{	ext{hyp}} = \frac{4}{5}$,the opposite side is $\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$.Thus,$	an x = \frac{	ext{opp}}{	ext{adj}} = \frac{3}{4}$,which implies $x = 	an ^{ - 1}\frac{3}{4}$.Now,the expression becomes $	an ^{ - 1}\frac{3}{4} + 	an ^{ - 1}\frac{3}{5}$.Using the formula $	an ^{ - 1}A + 	an ^{ - 1}B = 	an ^{ - 1}\left( \frac{A + B}{1 - AB} ight)$:$= 	an ^{ - 1}\left( \frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{3}{4} 	imes \frac{3}{5}} ight)$$= 	an ^{ - 1}\left( \frac{\frac{15 + 12}{20}}{1 - \frac{9}{20}} ight)$$= 	an ^{ - 1}\left( \frac{\frac{27}{20}}{\frac{11}{20}} ight)$$= 	an ^{ - 1}\frac{27}{11}$.

$\cos ^{ - 1}\frac{4}{5} + \tan ^{ - 1}\frac{3}{5} = $

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