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

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(B) Given that $	an A = \frac{1}{2}$ and $	an B = \frac{1}{3}$.We know the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substit

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$\sin 2B$

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(B) Given that $	an A = \frac{1}{2}$ and $	an B = \frac{1}{3}$.We know the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substituting the values: $	an(A + B) = \frac{1/2 + 1/3}{1 - (1/2)(1/3)} = \frac{5/6}{1 - 1/6} = \frac{5/6}{5/6} = 1$.Since $	an(A + B) = 1$,we have $A + B = 45^\circ$.Therefore,$2A = 90^\circ - 2B$.Taking cosine on both sides: $\cos 2A = \cos(90^\circ - 2B)$.Using the identity $\cos(90^\circ - 	heta) = \sin 	heta$,we get $\cos 2A = \sin 2B$.

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

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