2016/084) If $\cos(x-y) = \frac{4}{5}$ and $\sin(x+y) = \frac{5}{13}$ and x,y are between 0 and $\frac{\pi}{4}$ then find $\tan (2x)$

 we have $\cos(x-y) = \frac{4}{5}=> \sin (x-y) = \sqrt{1-(\cos(x-y))^2} = \sqrt{1-\frac{16}{25}} = \frac{3}{5}$
 and  $\sin(x+y) = \frac{5}{13}=> \cos (x+y) = \sqrt{1-(\sin(x+y))^2} = \sqrt{1-\frac{144}{169}} = \frac{12}{13}$

giving $\tan (x-y) = \frac{\sin (x-y}{\cos(x-y)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$
and $\tan (x+y) = \frac{\sin (x+y}{\cos(x+ y)} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$
hence
$\tan(2x) = \tan (x+y+x-y) = \frac{\tan (x+y) + \tan (x-y)}{1- \tan (x+y)\tan (x-y)} = \frac{\frac{5}{12} + \frac{3}{4}}{1- \frac{5}{12}\frac{3}{4}} = \frac{46}{33}$