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`

Options

A. `2npi, in Z`
B. `n pi/2, n in Z`
C. `(2n+1) pi/2, n in X`
D. `npi, n in Z`

Answer

C

Explanation

`AB=[(cos^(2) theta,),(cos theta sin theta,)][(cos^(2) phi,cos phi sin phi),(cos phi sin phi,sin^(2) phi)]`
`=[(cos^(2) theta cos^(2) phi+cos theta cos phi sin theta sin phi ,cos^(2)theta cos phi sin phi+cos theta sin theta sin^(2) phi),(cos theta sin thete cos^(2) phi+sin^(2) theta cos phi sin phi,cos theta cos phi sin theta sin phi+sin^(2) theta sin^(2) phi)]`
`=[(cos theta cos phi (cos (theta-phi)),cos theta sin phi (cos(theta-phi))),(sin theta cos phi (cos (theta-phi)),sin theta sin phi (cos (theta-phi)))]`
`=(cos (theta-phi)) [(cos theta cos phi,cos theta sin phi),(sin theta cos phi,sin theta sin phi)]`
Now, `AB=O implies cos (theta-phi)=0`
`implies theta-phi=(2n+1) pi//2, n in Z`.