The value of 2 tan ^{-1} frac{1}{2} + tan ^{- | vedclass

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(D) We know that $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.Applying this for $x = \frac{1}{2}$,we get $2 	an ^{-1} \frac{1}{2} =

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$	an ^{-1} \left( \frac{41}{12} ight)$

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(D) We know that $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.Applying this for $x = \frac{1}{2}$,we get $2 	an ^{-1} \frac{1}{2} = 	an ^{-1} \left( \frac{2 	imes (1/2)}{1 - (1/2)^2} ight) = 	an ^{-1} \left( \frac{1}{1 - 1/4} ight) = 	an ^{-1} \left( \frac{1}{3/4} ight) = 	an ^{-1} \left( \frac{4}{3} ight)$.Now,the expression becomes $	an ^{-1} \left( \frac{4}{3} ight) + 	an ^{-1} \left( \frac{3}{8} ight)$.Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$,we have:$	an ^{-1} \left( \frac{4/3 + 3/8}{1 - (4/3)(3/8)} ight) = 	an ^{-1} \left( \frac{(32+9)/24}{1 - 12/24} ight) = 	an ^{-1} \left( \frac{41/24}{12/24} ight) = 	an ^{-1} \left( \frac{41}{12} ight)$.

The value of $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{3}{8}$ is

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