" If "A=|[1,tan(theta)/(2)],[-tan(theta)/(2),1]|" and "AB=1," then "B

If A=[[1, tan((theta)/(2))], [-tan((theta)/(2)), 1]] and AB=I , then B=

If A=[(1,tan(theta/2)),(-tan(theta/2),1)] and AB=I , then B=

If A=[(1,tan(theta/2)),(-tan(theta/2),1)] and AB=I, then B= (A) {cos^2 (theta/2)}A (B) {cos^2(theta/2)}A\' (C) {cos^2(theta/2)}I (D) none of these

If A=[(1,tan(theta/2)),(-tan(theta/2),1)] and AB=I, then B= (A) {cos^2 (theta/2)}A (B) {cos^2(theta/2)}A\' (C) {cos^2(theta/2)}I (D) none of these

If A = [(1,tan""(theta)/(2)),(-tan""(theta)/(2),1)] and AB = I_(2) , then B=

If A = [(1,-tan""(theta)/(2)),(tan""(theta)/(2), 1)] and B = [(1,tan""(theta)/(2)),(-tan""(theta)/(2), 1)] show that, AB^(-1) = [(costheta,-sin theta),(sin theta, cos theta)] .

If A(theta)=[(1, than theta),(-tan, theta=1)] and AB=l , then (sec^(2)theta)B is equal to

[[1, -tan((theta)/(2))], [tan((theta)/(2)), 1]][[1, tan((theta)/(2))], [-tan((theta)/(2)), 1]]^(-1)=

Show that [[cos theta,-sin theta],[sin theta,cos theta]]=[[1,-tan(theta)/(2)],[tan(theta)/(2),1]][[1,tan(theta)/(2)],[-tan(theta)/(2),1]]^(-1)