If `A(cos theta, sin theta), B (sin theta, cos theta), C (1, 2)` are the vertices of `DeltaABC`. Find the locus of its centroid if `theta` varies.

sin^(3)theta + sin theta - sin theta cos^(2)theta =

If A=[[cos 2 theta, sin 2 theta],[-sin2theta, cos2theta]] , find A^2

(1+sin 2theta+cos 2theta)/(1+sin2 theta-cos 2 theta) =

(sin theta + sin 2 theta)/( 1 + cos theta + cos 2 theta) = tan theta.

(1+sin2theta+cos2theta)/(1+sin2theta-cos2theta)=?

If cos 3theta+sin3theta+(2sin2theta-3)(sin theta-cos theta)gt0 , then theta lies in

Prove : (cos theta)/(1+sin theta) + (cos theta)/(1-sin theta)= 2 sec theta

Let A(5, 12), B(-13 costheta, 13 sin theta) and C(13 sin theta, -13 cos theta) are angular points of ABC where theta in R . The locus of orthocentre of DeltaABC is

Prove that cos^6 theta+ sin^6 theta= 1-3sin^2 theta cos^2 theta

int (d theta)/((sin theta - 2 cos theta)(2 sin theta + cos theta))