The value of tan ^{-1}left(frac{1}{8}right)+t | vedclass

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(B) We use the formula $	an ^{-1}(x) + 	an ^{-1}(y) = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$.First,calculate $	an ^{-1}\left(\frac{1}{2}ight) +

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$\frac{\pi}{4}$

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(B) We use the formula $	an ^{-1}(x) + 	an ^{-1}(y) = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$.First,calculate $	an ^{-1}\left(\frac{1}{2}ight) + 	an ^{-1}\left(\frac{1}{5}ight)$:$	an ^{-1}\left(\frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \cdot \frac{1}{5}}ight) = 	an ^{-1}\left(\frac{\frac{7}{10}}{\frac{9}{10}}ight) = 	an ^{-1}\left(\frac{7}{9}ight)$.Now,add $	an ^{-1}\left(\frac{1}{8}ight)$ to the result:$	an ^{-1}\left(\frac{1}{8}ight) + 	an ^{-1}\left(\frac{7}{9}ight) = 	an ^{-1}\left(\frac{\frac{1}{8} + \frac{7}{9}}{1 - \frac{1}{8} \cdot \frac{7}{9}}ight)$.Simplify the expression inside the bracket:$\frac{\frac{9+56}{72}}{\frac{72-7}{72}} = \frac{65}{65} = 1$.Thus,the expression becomes $	an ^{-1}(1) = \frac{\pi}{4}$.

The value of $\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)$ is

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