The value of tan ^{-1} frac{1}{3}+tan ^{-1} f | vedclass

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(B) Let $L = 	an ^{-1} \frac{1}{3} + 	an ^{-1} \frac{1}{5} + 	an ^{-1} \frac{1}{7} + 	an ^{-1} \frac{1}{8}$.Using the formula $	an ^{-1} x + 	an

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$\frac{\pi}{4}$

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(B) Let $L = 	an ^{-1} \frac{1}{3} + 	an ^{-1} \frac{1}{5} + 	an ^{-1} \frac{1}{7} + 	an ^{-1} \frac{1}{8}$.Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$:First,group the terms: $L = \left( 	an ^{-1} \frac{1}{3} + 	an ^{-1} \frac{1}{5} ight) + \left( 	an ^{-1} \frac{1}{7} + 	an ^{-1} \frac{1}{8} ight)$.Calculate the first part: $	an ^{-1} \left( \frac{1/3 + 1/5}{1 - (1/3)(1/5)} ight) = 	an ^{-1} \left( \frac{8/15}{14/15} ight) = 	an ^{-1} \left( \frac{8}{14} ight) = 	an ^{-1} \left( \frac{4}{7} ight)$.Calculate the second part: $	an ^{-1} \left( \frac{1/7 + 1/8}{1 - (1/7)(1/8)} ight) = 	an ^{-1} \left( \frac{15/56}{55/56} ight) = 	an ^{-1} \left( \frac{15}{55} ight) = 	an ^{-1} \left( \frac{3}{11} ight)$.Now combine them: $L = 	an ^{-1} \left( \frac{4}{7} ight) + 	an ^{-1} \left( \frac{3}{11} ight) = 	an ^{-1} \left( \frac{4/7 + 3/11}{1 - (4/7)(3/11)} ight)$.$L = 	an ^{-1} \left( \frac{44/77 + 21/77}{1 - 12/77} ight) = 	an ^{-1} \left( \frac{65/77}{65/77} ight) = 	an ^{-1} (1) = \frac{\pi}{4}$.

The value of $\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ is $ . . . . . . $

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