Question : If $\tan (\alpha+\beta)=a, \tan (\alpha-\beta)=b$, then the value of $\tan 2 \alpha$ is:
Option 1: $\frac{a+b}{1-a b}$
Option 2: $\frac{a+b}{1+a b}$
Option 3: $\frac{a-b}{1+a b}$
Option 4: $\frac{a-b}{1-a b}$
Correct Answer: $\frac{a+b}{1-a b}$
Solution :
Given, $\tan (\alpha+\beta)=a$ and $\tan (\alpha-\beta)=b$
$\tan{2\alpha}=\tan((\alpha + \beta)+(\alpha - \beta)) = \frac{\tan(\alpha+\beta) + \tan(\alpha - \beta)}{1-\tan(\alpha + \beta)\tan(\alpha - \beta)}= \frac{a+b}{1-ab}$
Hence, the correct answer is $\frac{a+b}{1-ab}$.
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