If `theta-phi=pi/2`, then show that `[[cos^2theta,costhetasintheta],[costhetasintheta,sin^2theta]]*[[cos^2phi,cosphisinphi],[cosphisinphi,sin^2phi]]=0

We have
`AB=[{:(" "cos^(2)theta,costhetasintheta),(costhetasintheta," "sin^(2)theta):}][{:(cos^(2)phi,cosphisinphi),(cosphisinphi,sin^(2)phi):}]`
`[{:(cos^(2)thetacos^(2)phi+costhetasinthetacosphisinphi),(costhetasinthetacos^(2)phi+sin^(2)thetacosphisinphi):}`
`{:(cos^(2)thetacosphisinphi+costhetasinthetasin^(2)phi),(costhetasinthetacosphisinphi+sin^(2)thetasin^(2)phi):}]`
`=[{:(costhetacosphi.cos(theta-phi),costhetasinphi.cos(theta-phi)),(sinthetacosphi.cos(theta-phi),sinthetasinphi.cos(theta-phi)):}]`
`=[{:(0,0),(0,0):}]=[{:(because(theta-phi)" being an odd multiple of "(pi/2)","),(" we have cos "(theta-phi)=0):}].`