Search any question & find its solution
Question:
Answered & Verified by Expert
If $\sin 2 \theta$ and $\cos 2 \theta$ are solutions of $x^2+a x-c=0$, then
Options:
Solution:
2020 Upvotes
Verified Answer
The correct answer is:
$a^2+2 c-1=0$
$x^2+a x-c=0$
$\begin{aligned} & \sin 2 \theta, \cos 2 \theta \text { are roots } \\ & \sin 2 \theta+\cos 2 \theta=-a \\ & \sin 2 \theta \cdot \cos 2 \theta=-c \\ & (\sin 2 \theta+\cos 2 \theta)^2=\sin ^2 2 \theta+\cos ^2 2 \theta+2 \sin ^2 \theta \cos 2 \theta \\ & (-a)^2=1-2 c \\ & a^2=1-2 c \\ & a^2+2 c-1=0\end{aligned}$
$\begin{aligned} & \sin 2 \theta, \cos 2 \theta \text { are roots } \\ & \sin 2 \theta+\cos 2 \theta=-a \\ & \sin 2 \theta \cdot \cos 2 \theta=-c \\ & (\sin 2 \theta+\cos 2 \theta)^2=\sin ^2 2 \theta+\cos ^2 2 \theta+2 \sin ^2 \theta \cos 2 \theta \\ & (-a)^2=1-2 c \\ & a^2=1-2 c \\ & a^2+2 c-1=0\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.