Question
If $\text{a}\cos2\text{x}+\text{b}\sin2\text{x}=\text{c}$ has $\alpha$ and $\beta$ as its roots, then prove that, $\tan(\alpha+\beta)=\frac{\text{b}}{\text{a}}$

Answer

$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$ $\sin^2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$ substitute these valuse in the given equation, it reduces to $\text{a}(1-\tan^2)+\text{b}(2\tan\theta)=\text{c}(1+\tan^2\theta)$ $(\text{c+a})\tan^2\theta+2\text{b}\tan\theta+\text{C-a}=0$ As $\alpha$ and $\beta$ are roots $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{2\text{b}}{\text{c+a=c+a}}=\frac{\text{b}}{\text{a}}$

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