`z_2/z_1 =` (A) `e^(itheta) cos theta` (B) `e^(itheta) cos 2theta` (C) `e^(-itheta) cos theta` (D) `e^(2itheta) cos 2theta`

let z1, z2,z3 be vertices of triangle ABC in an anticlockwise order and angle ACB = theta then z_2-z_3 = CB/CA(z_1-z3) e^itheta .let p point on a circle with op diameter 2points Q & R taken on a circle such that angle POQ & QOR= theta if O be origin and PQR are complex no. z1, z2, z3 respectively then z_3^2/(z_1.z_2)= (A) sec^2theta.cos2theta (B) costheta.sec^2(2theta) (C) cos^2theta.sec2theta (D) sectheta.sec^2(2theta)

Solve cos theta+cos 2theta+cos 3theta=0 .

Prove: cos e c^2theta+sec^2theta=cos e c^2thetasec^2theta

Prove: (cos^2theta)/(sintheta)-cos e c\ theta+sintheta=0

(2 sin theta*cos theta - cos theta)/(1-sin theta+sin^2 theta-cos^2 theta) = cot theta

The value of sqrt((1+costheta)/(1-costheta)) is (a) cottheta-cos e c\ theta (b) cos e c\ theta+cottheta (c) cos e c^2theta+cot^2theta (d) (cottheta+cos e c\ theta)^2

Let theta in (pi//4,pi//2), then Statement I (cos theta)^(sin theta ) lt ( cos theta )^(cos theta ) lt ( sin theta )^(cos theta ) Statement II The equation e^(sin theta )-e^(-sin theta )=4 ha a unique solution.

The value of 3(cos theta-sin theta)^(4)+6(sin theta+cos theta)^(2)+4 sin^(6) theta is where theta in ((pi)/(4),(pi)/(2)) (a) 13-4cos^(4) theta (b) 13-4cos^(6) theta (c) 13-4cos^(6) theta+ 2 sin^(4) theta cos^(2) theta (d) 13-4cos^(4) theta+ 2 sin^(4) theta cos^(2) theta

c(1+cos2theta)= (A) c 2sin^2theta (B) c2cos^2theta (C) c2tan^2theta (D) none of these

If E(theta)=[[cos theta, sin theta] , [-sin theta, cos theta]] then E(alpha) E(beta)=