If $$2y \cos \theta = x \sin \theta and 2x \sec \theta - y \cosec \theta = 3$$, then the value of $$x^2 + 4y^2$$ is
$$let \theta = 45\degree$$
$$2y \frac{1}{\sqrt{2}} = x\frac{1}{\sqrt{2}} $$ = 2y= x
$$ 2x \sqrt{2} - y \sqrt{2} = 3$$
2x - y = $$\frac{3}{\sqrt{2}}$$ {substituting y= $$\frac{x}{2}$$}
x= $$Â \sqrt{2}$$
y= $$\frac{1}{Â \sqrt{2}}$$
value of $$x^2 + 4y^2$$ =Â $$\sqrt{2}^2 + 4\frac{1}{Â \sqrt{2}^2}$$ = 2+2 = 4
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