Trigonometry Related Problem involving Quadratic Equations

Question

If $\alpha$ and $\beta$ are the solutions of the equations $$\sin^2x+a\sin x+b=0$$ and $$\cos^2x+c \cos x+d=0$$ then $\sin(\alpha+\beta)=$? I cannot figure out how to start the sum and proceed with it.

Answer 1

Hint :

$$\begin{align} &\sin^2 \alpha + a\sin \alpha + b = \sin^2 \beta + a\sin \beta + b = 0 \\&\to \sin^2 \alpha - \sin^2 \beta +a\sin \alpha - a\sin \beta = 0 \\&\to (\sin \alpha - \sin \beta)(\sin \alpha + \sin \beta + a) = 0\end{align}$$

In the same manner :

$(\cos \alpha - \cos \beta)(\cos \alpha + \cos \beta + c) = 0$

Answer 2

$$\implies\sin\alpha+\sin\beta=-\dfrac a1$$

and $$\cos\alpha+\cos\beta=-\dfrac c1$$

On division and using Prosthaphaeresis Formulas, $$\dfrac ac=\tan\dfrac{\alpha+\beta}2$$ assuming $\cos\dfrac{\alpha-\beta}2\ne0$

Now $\sin2A=\dfrac{2\tan A}{1+\tan^2A}$