`alpha and beta` are the extremities of a focal chord of the ellipse, `x^2/a^2+y^2/b^2=1` then `(cos^2(alpha-beta)/2)/(cos^2(alpha+beta)/2)`

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 alpha , beta are eccentric angles of end points of a focal chord of the ellipse x^(2)/a^(2) + y^(2)/b^(2) =1 then tan(alpha /2) tan (beta/2) is equal to

If the eccentric angles of the extremities of a focal chord of an ellipse x^2/a^2 + y^2/b^2 = 1 are alpha and beta , then (A) e = (cos alpha + cos beta)/(cos (alpha + beta)) (B) e= (sin alpha + sin beta)/(sin(alpha + beta)) (C) cos((alpha-beta)/(2)) = e cos ((alpha + beta)/(2)) (D) tan alpha/2.tan beta/2 = (e-1)/(e+1)

If alpha,beta are the eccentric angles of the extremities of a focal chord of the ellipse x^(2)/16 + y^(2)/9 = 1 , then tan (alpha/2) tan(beta/2) =

Prove that the equation of the chord joining the points 'a'&'b' on the ellipse x^2/a^2+y^2/b^2=1 is x/a cos((alpha+beta)/2)+y/b sin((alpha+beta)/2)=cos((alpha-beta)/2)

cos^(2)(alpha+beta)+cos^(2)(alpha-beta)-cos2alpha cos2beta=

If cos(alpha-beta) and cos(alpha+beta) are the length of segments of a focal chord of parabola y^(2)=2(cos alpha)x divided at parabola cos alpha sec((beta)/(2)) is equal to

tan alpha and tan beta are roots of the equation x^(2)+ax+b=0, then the value of sin^(2)(alpha+beta)+a sin(alpha+beta)cos(alpha+beta)+b cos^(2)(alpha+beta) is equal to