2{tan ^{ - 1}}frac{1}{3} + {tan ^{ - 1}}frac{ | 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 $2	an ^{-1}\frac{1}{3}$: $2	an ^{-1}

Vedclass · Accepted Answer

$	an ^{ - 1}2$

Vedclass · Answer

(D) We use the formula $2	an ^{-1}x = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| Applying this to $2	an ^{-1}\frac{1}{3}$: $2	an ^{-1}\frac{1}{3} = 	an ^{-1}\left(\frac{2(1/3)}{1-(1/3)^2}ight) = 	an ^{-1}\left(\frac{2/3}{1-1/9}ight) = 	an ^{-1}\left(\frac{2/3}{8/9}ight) = 	an ^{-1}\left(\frac{2}{3} 	imes \frac{9}{8}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$. Now,the expression becomes $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{2}ight)$. Using the formula $	an ^{-1}x + 	an ^{-1}y = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$: $	an ^{-1}\left(\frac{3/4 + 1/2}{1 - (3/4)(1/2)}ight) = 	an ^{-1}\left(\frac{5/4}{1 - 3/8}ight) = 	an ^{-1}\left(\frac{5/4}{5/8}ight) = 	an ^{-1}\left(\frac{5}{4} 	imes \frac{8}{5}ight) = 	an ^{-1}(2)$. Thus,the correct option is $D$.

$2{\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{2} = $

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