# If $\cos(\alpha+\beta)=\large\frac{4}{5}$ and $\sin (\alpha-\beta)=\large\frac{5}{13}$ where $\alpha$ lies between $0$ and $\large\frac{\pi}{4}$,find the value of $\tan 2\alpha$

$\begin{array}{1 1}(A)\;\large\frac{56}{33}&(B)\;\large\frac{55}{36}\\(C)\;\large\frac{53}{36}&(D)\;\large\frac{56}{23}\end{array}$

## 1 Answer

Toolbox:
• $\tan A+\tan B=\large\frac{\tan A+\tan B}{1-\tan A.\tan B}$
Given :
$\cos(\alpha+\beta)=\large\frac{4}{5}$
$\sin (\alpha-\beta)=\large\frac{5}{13}$
$\tan 2\alpha=\tan(\alpha+\beta+\alpha-\beta)$
$\Rightarrow \large\frac{\tan(\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta).\tan(\alpha-\beta)}$
$\therefore \tan(\alpha+\beta)=\large\frac{3}{4}$
$\therefore \tan(\alpha-\beta)=\large\frac{5}{12}$
$\tan 2\alpha=\large\frac{\Large\frac{3}{4}+\frac{5}{12}}{1-\Large\frac{3}{4}\times \frac{5}{12}}$
$\Rightarrow \large\frac{14}{12}\times \frac{16}{11}=\frac{56}{33}$
Hence (A) is the correct answer.
answered Apr 18, 2014

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