$ an ^{-1}\left(\frac{1}{4} ight)+ an ^{-1}\left(\frac{2}{9} ight)$ is equal to

Question

$	an ^{-1}\left(\frac{1}{4}ight)+	an ^{-1}\left(\frac{2}{9}ight)$ is equal to

Aicharya · Accepted Answer

$	an ^{-1}\left(\frac{1}{4}ight)+	an ^{-1}\left(\frac{2}{9}ight)=	an ^{-1}\left\{\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} 	imes \frac{2}{9}}ight\}$ $=	an ^{-1}\left\{\frac{9+8}{36-2}ight\}=	an ^{-1}\left(\frac{1}{2}ight)$

$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)$ is equal to

Solution