`(sin theta + i cos theta)^2=?`.

((cos theta + i sin theta) / (sin theta + i cos theta)) ^ (4) = cos8 theta + i sin8 theta

2 cos (theta - pi //2) + 3 sin (theta + pi//2) - (3sin theta + 2 cos theta ) = ? A. cos theta - sin theta B. sin theta - cos theta C. sin theta + cos theta D. cot theta - tan theta .

[(1 + cos theta + i sin theta) / (sin theta + i (1 + cos theta))] ^ (4) = cos n theta + i sin n theta

If ((1 + cos theta + i sin theta ) /(i+ sin theta + i cos theta ))= cos ntheta + i sin n theta then n is equal to :

(sin theta-cos theta)^(2)*(sin theta+cos theta)^(2)=

Prove each of the following identities : (i) (sin theta - cos theta)/(sin theta + cos theta) + ( sin theta+ cos theta)/(sin theta - cos theta) = (2)/((2 sin^(2) theta -1)) (ii) (sin theta + cos theta ) /(sin theta - cos theta) + ( sin theta - cos theta) /(sin theta + cos theta) = (2) /((1- 2 cos^(2) theta))

In the form of A+iBz=(cos2 theta+i sin2 theta)^(-5)(cos3 theta-i sin3 theta)^(6)(sin theta-i cos theta)^(3) is