If tan A = -frac{1}{2} and tan B = -frac{1}{3 | vedclass

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(B) We are given $	an A = -\frac{1}{2}$ and $	an B = -\frac{1}{3}.$Using the formula for the tangent of the sum of two angles:$	an(A + B) = \frac{\

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

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(B) We are given $	an A = -\frac{1}{2}$ and $	an B = -\frac{1}{3}.$Using the formula for the tangent of the sum of two angles:$	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$Substituting the values:$	an(A + B) = \frac{-\frac{1}{2} - \frac{1}{3}}{1 - (-\frac{1}{2})(-\frac{1}{3})}$$	an(A + B) = \frac{-\frac{5}{6}}{1 - \frac{1}{6}} = \frac{-\frac{5}{6}}{\frac{5}{6}} = -1$Since $	an(A + B) = -1,$ and knowing that $	an(\frac{3\pi}{4}) = -1,$$A + B = \frac{3\pi}{4}.$

If $\tan A = -\frac{1}{2}$ and $\tan B = -\frac{1}{3},$ then $A + B = $

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