The numerical value of tan left(2 tan ^{-1}le | vedclass

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(C) Let $x = 2 	an ^{-1}\left(\frac{1}{5}ight)$.Then $	an \left(\frac{x}{2}ight) = \frac{1}{5}$.Using the formula $	an x = \frac{2 	an (x/2)}{

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$\frac{17}{7}$

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(C) Let $x = 2 	an ^{-1}\left(\frac{1}{5}ight)$.Then $	an \left(\frac{x}{2}ight) = \frac{1}{5}$.Using the formula $	an x = \frac{2 	an (x/2)}{1 - 	an^2 (x/2)}$,we get:$	an x = \frac{2(1/5)}{1 - (1/5)^2} = \frac{2/5}{1 - 1/25} = \frac{2/5}{24/25} = \frac{2}{5} 	imes \frac{25}{24} = \frac{5}{12}$.Now,we need to evaluate $	an \left(x + \frac{\pi}{4}ight)$.Using the formula $	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$	an \left(x + \frac{\pi}{4}ight) = \frac{	an x + 	an(\pi/4)}{1 - 	an x 	an(\pi/4)}$.Substituting $	an x = \frac{5}{12}$ and $	an(\pi/4) = 1$:$= \frac{5/12 + 1}{1 - (5/12)(1)} = \frac{17/12}{7/12} = \frac{17}{7}$.

The numerical value of $\tan \left(2 \tan ^{-1}\left(\frac{1}{5}\right)+\frac{\pi}{4}\right)$ is:

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