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

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(C) We use the formula for $	an ^{-1} x + 	an ^{-1} y$. Since $x 	imes y = 2 	imes 3 = 6 > 1$,the formula is: $	an ^{-1} x + 	an ^{-1} y = \pi +

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$\frac{3 \pi}{4}$

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

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

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