IF `alpha,beta` are the eccentric angles of the extremities of a focal chord of the ellipse `x^2/a^2+y^2/b^2=1`. Then show that `e cos""(alpha+beta)/2=cos""(alpha-beta)/2`

If alpha and beta are the eccentric angles of the ends of a focal chord of the ellipse then cos^(2)((alpha+beta)/(2))sec^(2)((alpha-beta)/(2)) =

If α, β are the eccentric angles of the extremeties of a focal chord of the ellipse (i) e cos (alpha+beta)/(2)=cos (alpha-beta)/(2) (ii) tan (alpha/2)tan (beta/2)=(e-1)/(e+1)

If alpha,beta are the ends of a focal chord of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 then its eccentricity e is

If alpha , beta are supplementary angles , then cos^(2)alpha + sin^(2) beta =

If alpha-beta is constant prove that the chord joining the points alpha and beta on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 touches a fixed ellipse

If alpha , beta are complementary angles , then sin^(2) alpha + sin^(2) beta =

If cos(theta - alpha)=a, cos(theta-beta)=b , then sin^(2)(alpha -beta) + 2abcos(alpha-beta)=0