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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & 2 	an ^{-1}\left(\frac{1}{3}ight)+\cos ^{-1}\left(\frac{3}{5}ight)=	an ^{-1} \frac{2 	imes \frac{1}{3}}{1-\left(\frac{1}{3}ight)^2}+\cos ^{-1}\left(\frac{3}{5}ight) \ & =	an ^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}ight)+\cos ^{-1}\left(\frac{3}{5}ight) \ & =	an ^{-1}\left(\frac{3}{4}ight)+\cot ^{-1}\left(\frac{3}{4}ight) \ & =\frac{\pi}{2}\end{aligned}$

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

Solution