The value of tan ^{-1}left(frac{1}{3}right)+t | vedclass

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(C) We use the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$.First,combine the first two terms:$	an ^{-1} \left(

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$\frac{\pi}{4}$

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(C) We use the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$.First,combine the first two terms:$	an ^{-1} \left( \frac{1}{3} ight) + 	an ^{-1} \left( \frac{1}{5} ight) = 	an ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} 	imes \frac{1}{5}} ight) = 	an ^{-1} \left( \frac{\frac{8}{15}}{\frac{14}{15}} ight) = 	an ^{-1} \left( \frac{8}{14} ight) = 	an ^{-1} \left( \frac{4}{7} ight)$.Now,combine the last two terms:$	an ^{-1} \left( \frac{1}{7} ight) + 	an ^{-1} \left( \frac{1}{8} ight) = 	an ^{-1} \left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} 	imes \frac{1}{8}} ight) = 	an ^{-1} \left( \frac{\frac{15}{56}}{\frac{55}{56}} ight) = 	an ^{-1} \left( \frac{15}{55} ight) = 	an ^{-1} \left( \frac{3}{11} ight)$.Finally,add the results:$	an ^{-1} \left( \frac{4}{7} ight) + 	an ^{-1} \left( \frac{3}{11} ight) = 	an ^{-1} \left( \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7} 	imes \frac{3}{11}} ight) = 	an ^{-1} \left( \frac{\frac{44+21}{77}}{\frac{77-12}{77}} ight) = 	an ^{-1} \left( \frac{65}{65} ight) = 	an ^{-1} (1) = \frac{\pi}{4}$.Thus,the correct option is $C$.

The value of $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ is

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