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If $\sin 2 \theta$ and $\cos 2 \theta$ are solutions of $x^2+b x-c=0$, then
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Verified Answer
The correct answer is:
$b^2+2 c-1=0$
According to given informations
$\sin 2 \theta+\cos 2 \theta=-b$ and $\sin 2 \theta \cos 2 \theta=-c$
$\because(\sin 2 \theta+\cos 2 \theta)^2=\sin ^2 2 \theta+\cos ^2 2 \theta+2 \sin 2 \theta \cos 2 \theta$
$\begin{aligned} & \Rightarrow(-b)^2=1+2(-c) \\ & \Rightarrow b^2+2 c-1=0\end{aligned}$
Hence, option (b) is correct.
$\sin 2 \theta+\cos 2 \theta=-b$ and $\sin 2 \theta \cos 2 \theta=-c$
$\because(\sin 2 \theta+\cos 2 \theta)^2=\sin ^2 2 \theta+\cos ^2 2 \theta+2 \sin 2 \theta \cos 2 \theta$
$\begin{aligned} & \Rightarrow(-b)^2=1+2(-c) \\ & \Rightarrow b^2+2 c-1=0\end{aligned}$
Hence, option (b) is correct.
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