If $\tan \frac{\alpha }{2},\tan \frac{\beta }{2}$ are the roots of the equation $8x^2-26x+15=0,$ then 

$$\begin{gathered} \tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{13}{4},\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{15}{8} \\ \Rightarrow \tan \frac{\alpha +\beta }{2}=\frac{\frac{13}{4}}{1-\frac{15}{8}}=\frac{-26}{7} \\ \Rightarrow \cos (\alpha +\beta )=\frac{1-{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}{1+{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}=\frac{1-\frac{676}{49}}{1+\frac{676}{49}}=\frac{-627}{725} \end{gathered}$$

    $$\begin{gathered} \sin (\alpha +\beta )=\frac{2\tan \frac{\alpha +\beta }{2}}{1-{\tan }^2\left(\frac{\alpha +\beta }{2}\right)} \\ =\frac{2(-26)}{7\left(1+\frac{676}{49}\right)}=\frac{-364}{725} \end{gathered}$$

   $$\begin{gathered} \tan (\alpha +\beta )=\frac{2\tan \frac{\alpha +\beta }{2}}{1-{\tan }^2\left(\frac{\alpha +\beta }{2}\right)} \\ =\frac{2(-26)}{7\left(1-\frac{676}{49}\right)}=\frac{364}{627} \end{gathered}$$