Question

If $\cos \left(\alpha +\beta \right)$ $=\frac{4}{5}$ $and$ $\sin (\alpha -\beta )$ $=\frac{5}{13}$ $\tan 2\alpha =?$

Answer 1

$\begin{array}{cc}\cos \left(\alpha +\beta \right)=\frac{4}{5}& \sin \left(\alpha -\beta \right)=\frac{5}{13}\\ \tan \left(\alpha +\beta \right)=\frac{3}{4}& \tan \left(\alpha -\beta \right)=\frac{5}{12}\\ \tan 2\alpha =\tan \left[\left(\alpha +\beta \right)+\left(\alpha -\beta \right)\right]& \\ =\frac{\tan \left(\alpha +\beta \right)+\tan \left(\alpha -\beta \right)}{1-\tan \left(\alpha +\beta \right).\tan \left(\alpha -\beta \right)}& \\ =\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}}=\frac{56}{33}& \end{array}$