" 13.If "A=[[cos2 theta,sin2 theta],[-sin2 theta,cos2 theta]]," find "A^(2)

If A=[[cos2 theta,sin2 theta],[-sin2 theta,cos2 theta]] then find A^2

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

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

(sin2 theta)/(1-cos2 theta)

cos theta+sin theta=cos2 theta+sin2 theta

(sin theta+sin2 theta)/(1+cos theta+cos2 theta)

(sin2 theta)/(1+cos2 theta)=

Inverse of the matrix [[cos2 theta,-sin2 thetasin2 theta,cos2 theta]] is

Inverse of the matrix [[cos2 theta,-sin2 thetasin2 theta,cos2 theta]] is

Let A = [ (cos theta, sin theta),(- sin theta, cos theta)] , then show that A ^(2) = [(cos 2 theta, sin 2 theta),( - sin 2 theta, cos 2 theta)]