If tan A = - frac{1}{2} and tan B = - frac{1} | 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 a sum: $	an(A + B) = \frac{	an A + 	an

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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 a sum: $	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})} = \frac{-\frac{5}{6}}{1 - \frac{1}{6}} = \frac{-\frac{5}{6}}{\frac{5}{6}} = -1$.Since $	an(A + B) = -1$,and considering the range of values for $A$ and $B$ in the second and fourth quadrants,we have $A + B = \frac{3\pi}{4}$.

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

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