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

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

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(A) We use the formula $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.For $x = \frac{1}{3}$,we have:$2 	an ^{-1} \left( \frac{1}{3} ight) = 	an ^{-1} \left( \frac{2 	imes \frac{1}{3}}{1 - (\frac{1}{3})^2} ight)$$= 	an ^{-1} \left( \frac{\frac{2}{3}}{1 - \frac{1}{9}} ight)$$= 	an ^{-1} \left( \frac{\frac{2}{3}}{\frac{8}{9}} ight)$$= 	an ^{-1} \left( \frac{2}{3} 	imes \frac{9}{8} ight) = 	an ^{-1} \left( \frac{3}{4} ight)$Now,substituting this back into the original expression:$2 	an ^{-1} \left( \frac{1}{3} ight) - 	an ^{-1} \left( \frac{3}{4} ight) = 	an ^{-1} \left( \frac{3}{4} ight) - 	an ^{-1} \left( \frac{3}{4} ight) = 0$.

$2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=$

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