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

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Question

(A) We use the formula $2 	an^{-1}(x) = 	an^{-1}\left(\frac{2x}{1-x^2}ight)$.For $x = \frac{1}{3}$,$2 	an^{-1}\left(\frac{1}{3}ight) = 	an^{-1

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$1$

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(A) We use the formula $2 	an^{-1}(x) = 	an^{-1}\left(\frac{2x}{1-x^2}ight)$.For $x = \frac{1}{3}$,$2 	an^{-1}\left(\frac{1}{3}ight) = 	an^{-1}\left(\frac{2(1/3)}{1-(1/3)^2}ight) = 	an^{-1}\left(\frac{2/3}{1-1/9}ight) = 	an^{-1}\left(\frac{2/3}{8/9}ight) = 	an^{-1}\left(\frac{2}{3} 	imes \frac{9}{8}ight) = 	an^{-1}\left(\frac{3}{4}ight)$.Now,the expression becomes $	an \left(	an^{-1}\left(\frac{3}{4}ight) + 	an^{-1}\left(\frac{1}{7}ight)ight)$.Using the formula $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$,we get:$	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)$.Finally,$	an(	an^{-1}(1)) = 1$.

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

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