If `A = [[cos^2theta, costhetasintheta],[costhetasintheta, sin^2theta]]` B= `[[cos^2phi, cosphisinphi], [cosphisinphi, sin^2phi]]` and `theta - phi = (2n+1)(pi)/2` Find AB.

Options

A. `theta=nphi,n=0,1,2,`. . . .
B. `theta+phi=npi,n=0,1,2` . . .
C. `theta=phi+(2n+1)(pi)/(2),n=0,1,2` . .
D. `theta=phi+n(pi)/(2),n=0,1,2,` . . .

Answer

C

Explanation

Now
`AB=[(cos^(2)theta,sinthetacostheta),(costhetasintheta,sin^(2)theta)]`
`[(cos^(2)theta,sinphicosphi),(cosphisinphi,sin^(2)phi)]`
`[(cos^(2)thetacos^(2)phi+sinthetacosphicosthetasinphicos^(2)thetasinphicosphi+sin^(2)phisinthetacostheta),(cos^(2)phicosthetasintheta+sin^(2)thetasinphicosphicosthetasinthetasinphicosphi+sin^(2)thetasin^(2)phi)]`
`=[(costhetacosphicos(theta-phi)sinphicosthetacos(theta-phi)),(sinthetacosphicos(theta-phi)sinthetasinphicos(theta-phi))]`
Since, AB=0
`impliescos(theta-phi)=0`
`impliescos(theta-phi)=cos(2n+1)(pi)/(2)`
`impliestheta=(2n+1)(pi)/(2)+phi`
where n=0,1,2, . . .