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

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(D) We use the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$.Given expression: $	an \left( 	an^{-1} \frac{1}{2

Vedclass · Accepted Answer

$1/7$

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(D) We use the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$.Given expression: $	an \left( 	an^{-1} \frac{1}{2} - 	an^{-1} \frac{1}{3} ight)$.Applying the formula with $x = \frac{1}{2}$ and $y = \frac{1}{3}$:$= 	an \left[ 	an^{-1} \left( \frac{\frac{1}{2} - \frac{1}{3}}{1 + (\frac{1}{2} 	imes \frac{1}{3})} ight) ight]$$= 	an \left[ 	an^{-1} \left( \frac{\frac{3-2}{6}}{1 + \frac{1}{6}} ight) ight]$$= 	an \left[ 	an^{-1} \left( \frac{1/6}{7/6} ight) ight]$$= 	an \left[ 	an^{-1} \left( \frac{1}{7} ight) ight]$$= \frac{1}{7}$.

The value of $\tan \left( \tan^{-1} \frac{1}{2} - \tan^{-1} \frac{1}{3} \right)$ is

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