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

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

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

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

If [1-tanthetatantheta1][1tantheta-tantheta1]^(-1)=[a-b b a] , then a=1,\ \ b=1 (b) a=cos2\ theta,\ \ b=sin2\ theta (c) a=sin2\ theta,\ \ b=cos2theta (d) none of these

If |(1,-t a ntheta),(t a ntheta,1)| |(1,t a ntheta),(-t a ntheta,1)|^-1=|[a,-b],[b ,a]| , then a. a=cos2theta b. a=1 c. b=sin2theta d. b=-1

If |(1,-t a ntheta),(t a ntheta,1)| |(1,t a ntheta),(-t a ntheta,1)|^-1=|[a,-b],[b ,a]| , then a. a=cos2theta b. a=1 c. b=sin2theta d. b=-1

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)]

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

If a tantheta=b , then a cos 2theta+b sin 2theta=