Statement 1 : `(a cos^2 theta, 2a cos theta)` is the parametric point of parabola `y^2 = 4ax`. Statement 2: For all value of `theta, -1 le cos theta le1`

1+cos2 theta+cos4 theta+cos6 theta=

cos theta cos2 theta cos3 theta=(1)/(4)

The value of (1+ cos theta)(1- cos theta) "cosec"^(2) theta =

For 0 le theta le pi/2 the maximum value of sin theta + cos theta is

Solve sin^(2) theta-cos theta=1/4, 0 le theta le 2pi .

2(cos theta+cos2 theta)+(1+2cos theta)sin2 theta=2sin theta

Solve for theta,2 cos^2 theta+sin theta-1=0,0 le theta le pi/2

Statement I If 2 cos theta + sin theta=1(theta != (pi)/(2)) then the value of 7 cos theta + 6 sin theta is 2. Statement II If cos 2theta-sin theta=1/2, 0 lt theta lt pi/2 , then sin theta+cos 6 theta = 0 .