If a cos2x+b sin2x=c has α and β as its roots,

Solution:

Given: $a \cos 2 x+b \sin 2 x=c$

$\Rightarrow a\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)+b\left(\frac{2 \tan x}{1+\tan ^{2} x}\right)-c=0$

$\Rightarrow a\left(1-\tan ^{2} x\right)+2 b \tan x-c\left(1+\tan ^{2} x\right)=0$

$\Rightarrow a-a \tan ^{2} x+2 b \tan x-c-c \tan ^{2} x=0$

$\Rightarrow(a+c) \tan ^{2} x-2 b \tan x+(c-a)=0$                ....(1)

This a quadratic equation in terms of $\tan ^{2} x$.

It is given that $\alpha$ and $\beta$ are the roots of the given equation, so $\tan \alpha$ and $\tan \beta$ are the roots of (1).

Since $\tan \alpha$ and $\tan \beta$ are the roots of the equation $(a+c) \tan ^{2} x-2 b \tan x+(c-a)=0 .$ Therefore,

(i)

$\tan \alpha+\tan \beta=-\frac{(-2 b)}{a+c} \quad\left(\right.$ Sum of roots $\left.=-\frac{b}{a}\right)$

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

(ii)

$\tan \alpha \tan \beta=\frac{c-a}{a+c} \quad$ (Product of roots $=\frac{c}{a}$ )

Or $\tan \alpha \tan \beta=\frac{c-a}{c+a}$

(iii)

$\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$

$=\frac{\frac{2 b}{a+c}}{1-\left(\frac{c-a}{c+a}\right)}$ [From (i) and (ii)]

$=\frac{\frac{2 b}{a+c}}{\frac{c+a-c+a}{c+a}}$

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

$=\frac{b}{a}$