If `A=[(cos^2 theta , sin theta cos theta ),(sin theta cos theta , sin^2 theta )], B=[(cos^2 phi , sin phi cos phi ),(sin phi cos phi , sin^2 phi )]` and `theta` and `phi` differs by `pi/2` , then AB=

Answer

N/a

Explanation

`L.H.S=[{:("cos"^(2)theta,costhetasintheta),(costheta sin theta,sin ^(2)theta):}][{:("cos"^(2)phi, cos phi sinphi),("cos" phisin phi ,sinphi):}]`
`=[{:(cos^(2)theta cos^(2)theta phi cosphi sintheta sinphi" "),(" "cos^(2)thetacosphisin +costhetasinthetasin^(2)phi),(costhetacos^(2)phisintheta+sin^(2)cosphi sinphi),(" "cos theta sin theta cosphi sinphi +sin^(2)thetasin^(2)phi):}]`
`=[{:(costheta cos phi( cos phi +sin theta sin phi )),(" "cos theta sin phi (costheta cos phi +sinthetasin phi)),(cosphisintheta(costhetacos phi +sintheta sin phi)),(" "sinthetasin phi (costhetacosphi +sintheta sinphi)):}]`
`=[{:(costheta cosphi .cos(theta-phi),costhetasin phi cos (theta-phi)),(cosphi sin theta.cos (theta-phi),sinthetasinphicos(theta-phi)):}]`
`=[{:(cos thetacos phi.cospi//2,costhetasinphi.cospi//2),(cosphisintheta.cospi//2,sinthetasinphi.cospi//2):}]`
`[{:(0,0),(0,0):}]=O=R.H.S.` hence proved