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If $\tan A=\frac{1}{2}, \tan B=\frac{1}{3}$ then $\tan (A+2 B)$ has the value
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$\tan A=\frac{1}{2}, \tan B=\frac{1}{3}$
[given] we have $\tan 2 B=\frac{2 \tan B}{1-\tan ^2 B}=\frac{3}{4}$
$\because \tan (A+2 B)=\frac{\tan A+\tan 2 B}{1-\tan A \cdot \tan 2 B}=\frac{\frac{1}{2}+\frac{3}{4}}{1-\frac{1}{2} \times \frac{3}{4}}=\frac{\frac{5}{4}}{\frac{5}{8}}=2$
[given] we have $\tan 2 B=\frac{2 \tan B}{1-\tan ^2 B}=\frac{3}{4}$
$\because \tan (A+2 B)=\frac{\tan A+\tan 2 B}{1-\tan A \cdot \tan 2 B}=\frac{\frac{1}{2}+\frac{3}{4}}{1-\frac{1}{2} \times \frac{3}{4}}=\frac{\frac{5}{4}}{\frac{5}{8}}=2$
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