If `x=a cos theta` and `y=bsin theta` so eliminate `theta`

Let x= cos theta and y = sin theta for any real value theta . Then x^(2)+y^(2)=

Find (dy)/(dx) if x= 3 cos theta - cos 2theta and y= sin theta - sin 2theta.

Find (dy)/(dx) if x=cos theta - cos 2 theta and" "y = sin theta - sin 2theta

If x/a cos theta + y/b sin theta = 1 , x/a sin theta - y/b cos theta = 1 then x^2/a^2 + y^2/b^2 =

If x = a cos theta + b sin theta and y =asin theta-b cos theta , then prove that y^2 (d^2y)/(dx^2)-x(dy)/(dx)+y=0

If x = a cos theta and y = bcot theta, show that : (a^(2))/(x^(2)) - (b^(2))/ (y^(2)) = 1

The normal to the curve x=a(cos theta + theta sin theta), y=a(sin theta - theta cos theta) at any theta is such that

If x = a( sin theta - theta cos theta) and y = a ( cos theta + theta sin theta) " find " (dy)/(dx) at theta = pi/4

Show that equations x=a cos theta + b sin theta, y = a sin theta - b cos theta represents a circle, wheter theta is a parameter.

If x cos theta + y sin theta =2 is perpendicular to the line x-y =3 then what is one of the value of theta ?