1.If 2ycosθ=xsinθ2y \cos \theta = x \sin \theta2ycosθ=xsinθ and 2xsecθ−ycosecθ=32x \sec \theta - y \cosec \theta = 32xsecθ−ycosecθ=3, then x2+4y2=x^2 + 4y^2 =x2+4y2=a.2b.4c.1d.None of theseLogin to continueOnly logged in users canattempt or see the solution.
