If sin 2 theta and cos 2 theta are solutions | vedclass

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(B) Given that $\sin 2 	heta$ and $\cos 2 	heta$ are the roots of the quadratic equation $x^2+bx-c=0$. From the properties of roots,the sum of roots

Vedclass · Accepted Answer

$b^2+2c-1=0$

Vedclass · Answer

(B) Given that $\sin 2 	heta$ and $\cos 2 	heta$ are the roots of the quadratic equation $x^2+bx-c=0$. From the properties of roots,the sum of roots is $\sin 2 	heta + \cos 2 	heta = -b$ and the product of roots is $\sin 2 	heta \cos 2 	heta = -c$. Squaring the sum of roots,we get $(\sin 2 	heta + \cos 2 	heta)^2 = (-b)^2$. Expanding this,$\sin^2 2 	heta + \cos^2 2 	heta + 2 \sin 2 	heta \cos 2 	heta = b^2$. Using the identity $\sin^2 A + \cos^2 A = 1$,we have $1 + 2(\sin 2 	heta \cos 2 	heta) = b^2$. Substituting the product of roots,$1 + 2(-c) = b^2$,which simplifies to $b^2 + 2c - 1 = 0$. Thus,option $B$ is correct.

If $\sin 2 \theta$ and $\cos 2 \theta$ are solutions of $x^2+bx-c=0$,then

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