[x=9sin20(1+cos0)],[y=-b cos20(1-cos^(2)theta)]

x=a sin2 theta(1+cos2 theta), y=b cos2 theta(1-cos2 theta) then (dy)/(dx)=

If x= a sin 2theta (1+ cos 2theta) , y= b cos 2theta(1- cos 2theta) , then (dy)/(dx) =

If x= a sin 2theta (1+ cos 2theta) , y= b cos 2theta(1- cos 2theta) , then (dy)/(dx) =

If x= a sin 2theta (1+ cos 2theta) , y= b cos 2theta(1- cos 2theta) , then (dy)/(dx) =

If =a sin2 theta(1+cos2 theta),y=b cos2 theta(1-os2 theta)x=a sin2 theta(1+cos2 theta),y=b cos2 theta(1-os2 theta) and (dy)/(dx)=k tan theta then the value of k is

If x=a sin2 theta(1+cos2 theta) and y=b cos2 theta(1-cos2 theta), then prove that (dy)/(dx)=(b)/(a)tan theta

(sin^(2) theta)/(1-cos theta)-(cos^(2) theta)/(1-sin theta)=cos theta-sin theta

If x = a Sin 2 theta (1+Cos2 theta) , y = b Cos2 theta (1-Cos2 theta) , then (dy)/(dx) =

A value of theta in (0, pi//3), for which |{:(1+"cos"^(2)theta, "sin"^(2)theta, 4"cos" 6 theta), ("cos"^(2)theta, 1+"sin"^(2)theta, 4"cos"6theta), ("cos"^(2)theta, "sin"^(2)theta, 1+4"cos"6theta):}| = 0, is