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

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(D) We know that $\cos ^{-1}\left(\frac{4}{5}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$ because in a right-angled triangle with base $4$ and hypote

Vedclass · Accepted Answer

$\frac{17}{6}$

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(D) We know that $\cos ^{-1}\left(\frac{4}{5}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$ because in a right-angled triangle with base $4$ and hypotenuse $5$,the perpendicular is $\sqrt{5^2 - 4^2} = 3$. Substituting this into the expression: $	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)$: $= 	an \left(	an ^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} 	imes \frac{2}{3}}ight)ight)$ $= 	an \left(	an ^{-1}\left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}ight)ight)$ $= 	an \left(	an ^{-1}\left(\frac{\frac{17}{12}}{\frac{6}{12}}ight)ight)$ $= 	an \left(	an ^{-1}\left(\frac{17}{6}ight)ight) = \frac{17}{6}$

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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