If $a\cos 2\theta+b\sin 2\theta=c$ has $\alpha$ and $\beta$ as its roots,then prove that $\tan\alpha+\tan \beta=\large\frac{2b}{a+c}$

Toolbox:
• $\cos 2\theta=\large\frac{1-\tan^2\theta}{1+\tan^2\theta}$
• $\sin2\theta=\large\frac{2\tan \theta}{1+\tan^2\theta}$
$\alpha$ and $\beta$ are the roots of equation
$a\cos 2\theta+b\sin 2\theta=c$
$a(\large\frac{1+\tan^2\theta}{1+\tan^2\theta})$$+b(\large\frac{2\tan\theta}{1+\tan^2\theta})$$=c$
$\large\frac{a(1+\tan^2\theta)+2b\tan\theta}{1+\tan^2\theta}$$=c a(1+\tan^2\theta)+2b\tan\theta=c(1+\tan^2\theta) \alpha and \beta are the roots a(1+\tan^2\alpha)+2b\tan\alpha=c(1+\tan^2\alpha)-----(1) a(1+\tan^2\beta)+2b\tan\beta=c(1+\tan^2\beta)-----(1) Subtract (2) from (1) a(\tan^2\beta-\tan^2\alpha)+2b(\tan\alpha-\tan\beta)=c(\tan^2\alpha-\tan^2\beta) \large\frac{a(\tan^2\beta-\tan^2\alpha)+2b(\tan\alpha-\tan\beta)}{(\tan\alpha-\tan \beta)}=$$c(\tan\alpha+\tan\beta)$
$-a(\tan \alpha+\tan \beta)+2b=c(\tan\alpha+\tan\beta)$
$2b=c(\tan\alpha+\tan \beta)+a(\tan \alpha+\tan \beta)$
$(\tan \alpha+\tan \beta)(a+c)=2b$
$\tan\alpha+\tan \beta=\large\frac{2b}{a+c}$
Hence proved.
Whole sol is wrong