1.If α,β\alpha,\betaα,β are eccentric angles of ends of a focal chord of x2+4y2=4x^2+4y^2=4x2+4y2=4, then 3cosα+β2=\sqrt{3}\cos\frac{\alpha+\beta}{2}=3cos2α+β=a.2cosα−β22\cos\frac{\alpha-\beta}{2}2cos2α−βb.2sinα−β22\sin\frac{\alpha-\beta}{2}2sin2α−βc.2secα+β22\sec\frac{\alpha+\beta}{2}2sec2α+βd.2sinα+β22\sin\frac{\alpha+\beta}{2}2sin2α+βLogin to continueOnly logged in users canattempt or see the solution.
