If 2tan A = 3tan B, then frac{sin 2B}{5 - cos | vedclass

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(B) Given: $2	an A = 3	an B \implies 	an A = \frac{3}{2}	an B$.Let $	an B = t$,then $	an A = \frac{3}{2}t$.We know that $\sin 2B = \frac{2t}{1 +

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$	an (A - B)$

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(B) Given: $2	an A = 3	an B \implies 	an A = \frac{3}{2}	an B$.Let $	an B = t$,then $	an A = \frac{3}{2}t$.We know that $\sin 2B = \frac{2t}{1 + t^2}$ and $\cos 2B = \frac{1 - t^2}{1 + t^2}$.Substituting these into the expression $\frac{\sin 2B}{5 - \cos 2B}$:$= \frac{\frac{2t}{1 + t^2}}{5 - \frac{1 - t^2}{1 + t^2}} = \frac{2t}{5(1 + t^2) - (1 - t^2)} = \frac{2t}{5 + 5t^2 - 1 + t^2} = \frac{2t}{4 + 6t^2} = \frac{t}{2 + 3t^2}$.Now,consider $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B} = \frac{\frac{3}{2}t - t}{1 + (\frac{3}{2}t)(t)} = \frac{\frac{1}{2}t}{1 + \frac{3}{2}t^2} = \frac{t}{2 + 3t^2}$.Since both expressions are equal,the result is $	an(A - B)$.

If $2\tan A = 3\tan B,$ then $\frac{\sin 2B}{5 - \cos 2B}$ is equal to

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