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

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(A) Given: $	an A = \frac{1}{2}$ and $	an B = \frac{1}{3}$.First,calculate $	an 2B$ using the formula $	an 2B = \frac{2 	an B}{1 - 	an^2 B}$.$

Vedclass · Accepted Answer

$2$

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(A) Given: $	an A = \frac{1}{2}$ and $	an B = \frac{1}{3}$.First,calculate $	an 2B$ using the formula $	an 2B = \frac{2 	an B}{1 - 	an^2 B}$.$	an 2B = \frac{2(\frac{1}{3})}{1 - (\frac{1}{3})^2} = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} 	imes \frac{9}{8} = \frac{3}{4}$.Now,use the formula $	an (A + 2B) = \frac{	an A + 	an 2B}{1 - 	an A \cdot 	an 2B}$.$	an (A + 2B) = \frac{\frac{1}{2} + \frac{3}{4}}{1 - (\frac{1}{2} 	imes \frac{3}{4})} = \frac{\frac{2+3}{4}}{1 - \frac{3}{8}} = \frac{\frac{5}{4}}{\frac{5}{8}} = \frac{5}{4} 	imes \frac{8}{5} = 2$.

If $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3}$,then the value of $\tan (A + 2B)$ is:

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