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If theta-phi=pi/2, prove that, [(cos^2 t...

If `theta-phi=pi/2,` prove that, `[(cos^2 theta,cos theta sin theta),(cos theta sin theta,sin^2 theta)] [(cos^2 phi,cos phi sin phi),(cos phi sin phi,sin^2 phi)]=0`

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`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
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