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$\sin \theta$ and $\cos\theta$ are roots of equation $ax^2+bx+c=0$ then,

$\begin{array}{1 1}(A)\;a^2-b^2+2ac=0&(B)\;a^2-b^2-2ac=0\\(C)\;a^2+b^2-2ac=0&(D)\;a^2+b^2+2ac=0\end{array} $

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$(\sin \theta+\cos\theta)^2=\sin^2\theta+\cos^2\theta+2\sin \theta\cos\theta$
$\Rightarrow 1+2\sin \theta\cos\theta$
$\Rightarrow (-\large\frac{b}{a})^2=$$1+\large\frac{2c}{a}$
$\Rightarrow a^2-b^2+2ac=0$
Hence (A) is the correct answer.
answered Apr 15, 2014 by sreemathi.v
 

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