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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & 	an \left[\cos ^{-1}\left(\frac{4}{5}ight)+	an ^{-1}\left(\frac{2}{3}ight)ight] \ & =	an \left[	an ^{-1}\left(\frac{3}{4}ight)+	an ^{-1}\left(\frac{2}{3}ight)ight]=	an \left[	an ^{-1}\left(\frac{\left(\frac{3}{4}ight)+\left(\frac{2}{3}ight)}{1-\left(\frac{3}{4}ight)+\left(\frac{2}{3}ight)}ight)ight] \ & =	an \left[	an ^{-1}\left(\frac{17}{6}ight)ight]=\frac{17}{6}\end{aligned}$

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

Solution