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

If AB=O for the matrices A=[[cos^2theta,costhetasintheta],[costhetasintheta,sin^2theta]] and B=[[cos^2phi,cosphisinphi],[cosphisinphi,sin^2phi]] then theta-phi is

Select and write the correct answer from the given alternatives in each of the following:If AB=0 where A=[[cos^2theta,costhetasintheta],[costhetasintheta,sin^2theta]] and B=[[cos^2phi,cosphisinphi],[cosphisinphi,sin^2phi]] ,then (theta-phi) is equal to

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

Flind the product of two matrices A =[[cos^(2) theta , cos theta sin theta],[cos theta sin theta ,sin^(2)theta]] B= [[cos^(2) phi,cos phi sin phi],[cos phisin phi,sin^(2)phi]] Show that, AB is the zero matrix if theta and phi differ by an odd multipl of pi/2 .

Flind the product of two matrices A =[[cos^(2) theta , cos theta sin theta],[cos theta sin theta ,sin^(2)theta]] B= [[cos^(2) phi,cos phi sin phi],[cos phisin phi,sin^(2)phi]] Show that, AB is the zero matrix if theta and phi differ by an odd multipl of pi/2 .

If A=[(cos^2theta,costheta sintheta),(cos theta sintheta, sin^2theta)] and B=[(cos^2 phi, cos phi sin phi),(cos phisinphi,sin^2phi)], show that AB is zero matrix if theta and phi differ by an odd multiple of pi/2.

The product of matrices A = [(cos^(2) theta, cos theta sin theta),(cos theta sin theta , sin^(2) theta)] and sin B = [(cos^(2)phi, cos phi sin phi),(cos phi sin phi, sin^(2) phi)] is a null matrix if theta - phi =