If  $\alpha ,\beta$are different values of  $x$satisfying  $a\cos x+b\sin x=c$, then  $\tan \left(\frac{\alpha +\beta }{2}\right)=$

$a\cos x+b\sin x=c$

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

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

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

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

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