prove that the equation of straight line which passes through the point (a cos^3a, sin^3a) & is perpendicular to the straight line x sec a+y cosec a=0 is x cos a-y sin a = a cos 2a.
The line x sec a + y csc a = 0
has slope -tan a.
So, we want a line with slope cot a
The line is thus
y - sin^3a = cota (x - acos^3a)
y sina - sin^4a = x cosa - a cos^4a
x cosa - y sina = a cos^4a - sin^4a
x cosa - y sina = a(cos^2a + sin^2a)(cos^2a - sin^2a)
x cosa - y sina = a cos2a