1.Eccentricity of x2−y2cosec2α=25x^2 - y^2\cosec^2\alpha = 25x2−y2cosec2α=25 is 5\sqrt{5}5 times that of x2cosec2α+y2=5x^2\cosec^2\alpha + y^2 = 5x2cosec2α+y2=5. Then α\alphaα isa.tan−12\tan^{-1}\sqrt{2}tan−12b.sin−13/4\sin^{-1}\sqrt{3/4}sin−13/4c.tan−12/5\tan^{-1}\sqrt{2/5}tan−12/5d.sin−12/5\sin^{-1}\sqrt{2/5}sin−12/5Login to continueOnly logged in users canattempt or see the solution.
