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If $\cos (\alpha+\beta)=\frac{4}{5}, \sin (\alpha-\beta)=\frac{5}{13}$ and $\alpha, \beta$ between 0 and $\frac{\pi}{4}$, then $\tan 2 \alpha$ is equal to
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Verified Answer
The correct answer is:
$\frac{56}{33}$
Given that
$$
\begin{aligned}
& \cos (\alpha+\beta)=\frac{4}{5} \Rightarrow \tan (\alpha+\beta)=\frac{3}{4} \\
& \sin (\alpha-\beta)=\frac{5}{13} \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \\
& \text { Now, } \tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)] \\
& =\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)} \\
& =\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}=\frac{(9+5) 4}{48-15}=\frac{56}{33} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \cos (\alpha+\beta)=\frac{4}{5} \Rightarrow \tan (\alpha+\beta)=\frac{3}{4} \\
& \sin (\alpha-\beta)=\frac{5}{13} \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \\
& \text { Now, } \tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)] \\
& =\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)} \\
& =\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}=\frac{(9+5) 4}{48-15}=\frac{56}{33} \\
&
\end{aligned}
$$
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