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

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 .

The product of matrices A=[[cos^2theta,sinthetacostheta],[sinthetacostheta,sin^2theta]] and B=[[cos^2phi,sinphicosphi],[sinphicosphi,sin^2phi]] is a null matrix if theta-phi= (A) 2npi,n in Z (B) (npi)/2, n in Z (C) (2n+1)pi/2, n in Z (D) npi, n in Z

If A=[[cos 2 theta, sin 2 theta],[-sin2theta, cos2theta]] , find A^2

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

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

If A=[[cos theta, sin theta],[sin theta, cos theta]],B=[[cos phi, sin phi],[sin phi, cos phi]] show that AB=BA.

(1+sin2theta+cos2theta)/(1+sin2theta-cos2theta)=?

Prove : (1+sin2theta-cos2theta)/(1+sin2theta+cos2theta)=tantheta

cos2thetacos2phi+sin^2(theta-phi)-sin^2(theta+phi)=

If f (theta) = [[cos^(2) theta , cos theta sin theta,-sin theta],[cos theta sin theta , sin^(2) theta , cos theta ],[sin theta ,-cos theta , 0]] ,then f ( pi / 7) is