If in a triangle ABC,A = tan^{-1} 2 and B = t | vedclass

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(C) Given that $A = 	an^{-1} 2$ and $B = 	an^{-1} 3$. We know that in a triangle $ABC$,$A + B + C = \pi$. Substituting the values,we get $	an^{-1}

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$\pi / 4$

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(C) Given that $A = 	an^{-1} 2$ and $B = 	an^{-1} 3$. We know that in a triangle $ABC$,$A + B + C = \pi$. Substituting the values,we get $	an^{-1} 2 + 	an^{-1} 3 + C = \pi$. Using the formula $	an^{-1} x + 	an^{-1} y = \pi + 	an^{-1} \left( \frac{x+y}{1-xy} ight)$ for $xy > 1$,we have: $	an^{-1} 2 + 	an^{-1} 3 = \pi + 	an^{-1} \left( \frac{2+3}{1-2 	imes 3} ight) = \pi + 	an^{-1} \left( \frac{5}{-5} ight) = \pi + 	an^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Now,$\frac{3\pi}{4} + C = \pi$. Therefore,$C = \pi - \frac{3\pi}{4} = \frac{\pi}{4}$.

If in a triangle $ABC$,$A = \tan^{-1} 2$ and $B = \tan^{-1} 3$,then angle $C$ is equal to

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