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

If [(1,-tan theta),(tan theta,1)][(1,tan theta),(-tan theta,1)]^(-1)=[(a,-b),(b,a)], then

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=[[0,-tan(theta/2)],[tan(theta/2),0]](theta!=npi,n in Z),B=[[costheta,-sintheta],[sintheta,costheta]] and I=[[1,0],[0,1]], then the value of (2I)/(costheta+1) is equal to

For 0

(a+b)tan(theta-phi)=(a-b)tan(theta+phi) then (sin 2theta)/(sin 2phi)=

If theta_(1) and theta_(2) are two values lying in [0,2 pi] for which tan theta=lambda, then tan((theta_(1))/(2))tan((theta_(2))/(2)) is equal to (a)0(b)-1(c)2(d)1