" 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 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

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)

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)] .

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

A = [(0,-tan""(theta)/(2)),(tan""(theta)/(2),0)] and (I+A)(I-A)^-1=[(a,-b),(b,a)] . Find 13(a^2+b^2)

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

(tan^2theta)/(1+tan^2theta) = ……………..