Simplify `cos theta[(cos theta,sin theta),(-sin theta,cos theta)]+sin theta[(sin theta,-cos theta),(cos theta,sin theta)]`

[[ Simplify :cos thetacos theta,sin theta-sin theta,cos theta]]+sin theta[[sin theta,-cos thetacos theta,sin theta]]

Simplify, costheta[[cos theta, sin theta],[-sin theta, cos theta]] + sin theta [[sin theta, -cos theta],[cos theta, sin theta]]

(i) |(cos theta,-sin theta),(sin theta,cos theta)|

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

If A = [(cos theta, sin theta),(-sin theta, cos theta)] and B = [(sin theta, - cos theta), (cos theta, sin theta)] , evaluate A cos theta + B sin theta .

cos theta+sin theta=cos2 theta+sin2 theta

If A=[(-cos theta, sin theta),(sin theta, cos theta)] then IAI=

If f(theta)=|(cos^(2) theta, cos theta sin theta, -sin theta),(cos theta sin theta, sin^(2) theta, cos theta),(sin theta, -cos theta, 0)| then f((pi)/12)=