If $\cot \frac{\alpha }{2},\cot \frac{\beta }{2}$ are roots of the equation $15x^2-26x+8=0,\text{ then }\cos (\alpha +\beta )$ is

 $\because \cot \frac{\alpha }{2},\cot \frac{\beta }{2}\text{ are roots of }15x^2-26x+8=0$

Then,  $\cot \frac{\alpha }{2}\cot \frac{\beta }{2}=\frac{8}{15}$

          $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{15}{8} ...\left(i\right)$

and   $\cot \frac{\alpha }{2}+\cot \frac{\beta }{2}=\frac{26}{15}$

        $\Rightarrow \frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{\tan \frac{\alpha }{2}\cdot \tan \frac{\beta }{2}}=\frac{26}{15}$

$\begin{array}{l}\Rightarrow \frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{\frac{15}{8}}=\frac{26}{15}\\ \therefore \tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{26}{15}\times \frac{15}{8}\\ \tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{26}{8} ...\left(ii\right)\end{array}$

We know that

  $\tan \left(\frac{\alpha }{2}+\frac{\beta }{2}\right)=\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\cdot \tan \frac{\beta }{2}}=\frac{\frac{26}{8}}{1-\frac{15}{8}}=\frac{\frac{26}{8}}{\frac{-7}{8}}$ [from Eqs. (i) and (ii) ]

  $\tan \left(\frac{\alpha +\beta }{2}\right)=-\frac{26}{7} ...\left(iii\right)$

and $\cos (\alpha +\beta )=\frac{1-{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}{1+{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}$

      $=\frac{1-{\left(-\frac{26}{7}\right)}^2}{1+{\left(-\frac{26}{7}\right)}^2}=\frac{1-\frac{676}{49}}{1+\frac{676}{49}}=-\frac{627}{725}$