2 tan ^{-1}left(frac{1}{3}right)+tan ^{-1}lef | vedclass

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(D) We use the formula $2 	an ^{-1}(x) = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| < 1$. Applying this to the first term with $x = \frac{1}{3

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$\frac{\pi}{4}$

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(D) We use the formula $2 	an ^{-1}(x) = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| Applying this to the first term with $x = \frac{1}{3}$: $2 	an ^{-1}\left(\frac{1}{3}ight) = 	an ^{-1}\left(\frac{2(\frac{1}{3})}{1-(\frac{1}{3})^2}ight) = 	an ^{-1}\left(\frac{2/3}{1-1/9}ight) = 	an ^{-1}\left(\frac{2/3}{8/9}ight) = 	an ^{-1}\left(\frac{2}{3} 	imes \frac{9}{8}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$. Now, the expression becomes $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{7}ight)$. Using the formula $	an ^{-1}(x) + 	an ^{-1}(y) = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$ where $xy $	an ^{-1}\left(\frac{3/4 + 1/7}{1 - (3/4)(1/7)}ight) = 	an ^{-1}\left(\frac{(21+4)/28}{1 - 3/28}ight) = 	an ^{-1}\left(\frac{25/28}{25/28}ight) = 	an ^{-1}(1) = \frac{\pi}{4}$.

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ is equal to

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