tan^{-1} frac{1}{2} + tan^{-1} frac{1}{3} = ? | vedclass

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(B) We use the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x + y}{1 - xy} ight)$ for $xy < 1$.Given $x = \frac{1}{2}$ and $y = \frac

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$\pi / 4$

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(B) We use the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x + y}{1 - xy} ight)$ for $xy Given $x = \frac{1}{2}$ and $y = \frac{1}{3}$,we have $xy = \frac{1}{6} Substituting the values:$	an^{-1} \frac{1}{2} + 	an^{-1} \frac{1}{3} = 	an^{-1} \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} 	imes \frac{1}{3}} ight)$$= 	an^{-1} \left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} ight)$$= 	an^{-1} \left( \frac{5/6}{5/6} ight)$$= 	an^{-1} (1)$$= \frac{\pi}{4}$.

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

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