If `A ={ theta : 2cos^2 theta + sintheta <=2}` , and `B = {theta: pi/2<=theta<= 3pi/2}` , then the region for `(AnnB)` is

A ={ theta : 2cos^2 theta + sintheta <=2} , and B = {theta: pi/2<=theta<= (3pi)/2} then (A∩B) is

The number of solutions of the paires of equations : 2sin^(2)theta - cos2theta = 0 , 2cos^(2)theta - 3sintheta = 0 in the interval [0,2pi] is :

If A = [(sintheta ,- costheta), (costheta, sintheta)] and B = [(costheta , sin theta),(-sintheta, cos theta)] . Compute (sin theta) A + (cos theta) B.

Find the acute angle theta such that 2cos^2theta=3sintheta .

2-cos^(2)theta =3 sintheta costheta , sintheta ne cos theta , then the value of tan theta is

Minimum value of the expression cos^2theta-(6 sintheta cos theta) + 3 sin^2theta + 2 , is

Prove : (cos^2theta)/(1-tan theta) + (sin^3theta)/(sintheta - cos theta) = 1 + sin theta cos theta

The value of (2(sin2theta+2cos^2theta-1))/(costheta-sintheta-cos3theta+sin3theta) is adot costheta b. sectheta c. cosectheta d. sintheta

The value of (2(sin2theta+2cos^2theta-1))/(costheta-sintheta-cos3theta+sin3theta) is a.\ costheta b. sectheta c. cosectheta d. sintheta