tan ^{-1} frac{1}{3}+tan ^{-1} frac{1}{5}+tan | vedclass

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(A) We use the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$.First,group the terms:$\left( 	an ^{-1} \frac{1}{3}

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$\frac{\pi}{4}$

Vedclass · Answer

(A) We use the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$.First,group the terms:$\left( 	an ^{-1} \frac{1}{3} + 	an ^{-1} \frac{1}{5} ight) + \left( 	an ^{-1} \frac{1}{7} + 	an ^{-1} \frac{1}{8} ight)$Apply the formula to the first pair:$	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)$Apply the formula to the second pair:$	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)$Now add the results:$	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}$

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ has the value

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