If `((1,-tantheta), (tantheta,1))((1,tantheta), (-tantheta,1))^(-1)=[(a,-b), (b,a)],` then

If A=[(1,-tantheta),(tantheta, 1)] [(1,tantheta),(-tantheta,1)]^-1 = [(a,-b),(b,a)] then (A) a=b=-1 (B) a=sin2theta, b=cos2theta (C) a=cos2theta, b=sin2theta (D) none of these

(1-tantheta)(1+sin 2theta) = (1+tantheta)

(tantheta+2)(2 tantheta+1)= ?

Solve: (1-tantheta)(1+sin2theta)=(1+tantheta)

Solve: (1-tantheta)(1+sin2theta)=(1+tantheta)

costheta(tantheta+2)(2tantheta+1)=?

((1-tantheta)/(1-cottheta))^(2)+1=

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

(tantheta+2)(2 tantheta+1) = 5 tantheta+ sec^(2)theta