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If $\alpha $ and $\beta $ are solutions of $sin^2\,x + a\, sin\, x + b = 0$ as well that of $cos^2\,x + c\, cos\, x + d = 0$ , then $sin\,(\alpha + \beta )$ is equal to
$\frac{{2bd}}{{{b^2} + {d^2}}}$
$\frac{{{a^2} + {c^2}}}{{2ac}}$
$\frac{{{b^2} + {d^2}}}{{2bd}}$
$\frac{{2ac}}{{{a^2} + {c^2}}}$
Solution
According to the given condition. $\sin \alpha+\sin \beta=-a$ and $\cos \alpha+\cos \beta=-c$
$\Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=-\mathrm{a}$ …..$(1)$
$\text { and } 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=-\mathrm{c}$ …..$(2)$
$(1) \div(2)$
$\Rightarrow \tan \frac{\alpha+\beta}{2}=\frac{a}{c}$
$\Rightarrow \sin (\alpha+\beta)=\frac{2 \tan \frac{\alpha+\beta}{2}}{1+\tan ^{2} \frac{\alpha+\beta}{2}}=\frac{2 a c}{a^{2}+c^{2}}$
Hence, $( 4)$ is the correct answer.
