If x + y = 3 - cos 4theta and x - y = 4 sin 2 | vedclass

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(D) Given equations are:$x + y = 3 - \cos 4	heta = 3 - (1 - 2 \sin^2 2	heta) = 2 + 2 \sin^2 2	heta$$x - y = 4 \sin 2	heta$Adding the two equations

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$\sqrt{x} + \sqrt{y} = 2$

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(D) Given equations are:$x + y = 3 - \cos 4	heta = 3 - (1 - 2 \sin^2 2	heta) = 2 + 2 \sin^2 2	heta$$x - y = 4 \sin 2	heta$Adding the two equations:$2x = 2 + 2 \sin^2 2	heta + 4 \sin 2	heta = 2(1 + 2 \sin 2	heta + \sin^2 2	heta) = 2(1 + \sin 2	heta)^2$$x = (1 + \sin 2	heta)^2 \implies \sqrt{x} = 1 + \sin 2	heta$Subtracting the two equations:$2y = 2 + 2 \sin^2 2	heta - 4 \sin 2	heta = 2(1 - 2 \sin 2	heta + \sin^2 2	heta) = 2(1 - \sin 2	heta)^2$$y = (1 - \sin 2	heta)^2 \implies \sqrt{y} = 1 - \sin 2	heta$Adding $\sqrt{x}$ and $\sqrt{y}$:$\sqrt{x} + \sqrt{y} = (1 + \sin 2	heta) + (1 - \sin 2	heta) = 2$Thus,the correct option is $D$.

If $x + y = 3 - \cos 4\theta$ and $x - y = 4 \sin 2\theta$,then:

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