If $\alpha ,\beta$are two values of $\theta$obtained from the equation  $a\cos \theta +b\sin \theta =c$. Then the value of $\tan \left(\frac{\alpha +\beta }{2}\right)$ is ____

$a\left(\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)+b\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=c$

$\Rightarrow a-a{\tan }^2\frac{\theta }{2}+2b\tan \frac{\theta }{2}=c+c{\tan }^2\frac{\theta }{2}$

$\Rightarrow (c+a){\tan }^2\frac{\theta }{2}-2b\tan \frac{\theta }{2}+c-a=0\left(i\right)$

$Let\tan \frac{\alpha }{2},\tan \frac{\beta }{2}aretherootsof\left(i\right)$

$\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{2b}{c+a},\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{c-a}{c+a}$

$\tan \left(\frac{\alpha +\beta }{2}\right)=\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=\frac{\frac{2b}{c+a}}{1-\left(\frac{c-a}{c+a}\right)}$

$=\frac{2b}{2a}=\frac{b}{a}$