If `((1,-tantheta), (tantheta,1))((1,tantheta), (-tantheta,1))^(-1)=[(a,-b), (b,a)],` then (i) `a=cos2theta` (ii)`b=1` (iii)`b=sin2theta` (iv)`b=-1`

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

If [(1,-tantheta),(tantheta,1)][(1,tantheta),(-tantheta,1)]^(-1)=[(a,-b),(b, a)] , then a=1,\ \ b=0 (b) a=cos2\ theta,\ \ b=sin2\ theta (c) a=sin2\ theta,\ \ b=cos2theta (d) none of these

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

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

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

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

Select the correct option from the given alternatives: If [[1,-tantheta],[tantheta,1]]([[1,tantheta],[-tantheta,1]])^-1 = [[a,-b],[b,a]] then, i] a=1,b=1 ii] a=cos2theta,b=sin2theta iii] a=sin2theta,b=cos2theta iv] a=0,b=1

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

Prove that (sin2theta)/(1+cos2theta)=tantheta

Prove that (1+sin2theta)/(1-sin2theta)=((1+tantheta)/(1-tan theta))^(2)