We have $a\sec\theta+b\tan\theta=c$-----(1)
$a\sec\theta=c-b\tan\theta$
Squaring on both sides
$a^2\sec^2\theta=c^2+b^2\tan^2\theta-2bc\tan\theta$
$\sec^2\theta=1+\tan^2\theta$
$a^2(1+\tan^2\theta)=c^2+b^2\tan^2\theta-2bc\tan\theta$
$a^2+a^2\tan^2\theta-b^2\tan^2\theta+2bc\tan\theta-c^2=0$
$(a^2-b^2)\tan^2\theta+2bc\tan\theta+(a^2-c^2)=0$-----(2)
It is given that $\alpha$ and $\beta$ are the solutions of equation (1)
$\therefore\tan\alpha$ and $\tan \beta$ will be roots of the quadratic equation (2)
$\tan\alpha+\tan \beta=-\large\frac{2bc}{a^2-b^2}$
$\tan\alpha\tan \beta=-\large\frac{a^2-c^2}{a^2-b^2}$
$\therefore \tan(\alpha+\beta)=\large\frac{\tan \alpha+\tan \beta}{1-\tan \alpha\tan \beta}$
$\Rightarrow \large\frac{-\Large\frac{2bc}{a^2-b^2}}{1-\Large\frac{a^2-c^2}{a^2-b^2}}$
$\Rightarrow \large\frac{-2bc}{c^2-b^2}\times \frac{a^2-b^2}{a^2-b^2-a^2+c^2}$
$\Rightarrow \large\frac{-2bc}{c^2-b^2}= \frac{2bc}{b^2-c^2}$
Hence proved