The region of y for which `y+cos theta=sin theta`has a real solution is `-sqrt(2)<=y<=sqrt(2)` `y>sqrt(2)` `y<=-sqrt(2)` `y<=1`

If y + "cos" theta = "sin" theta has a real solution, then

If the equation "sin" theta ("sin" theta + 2 "cos" theta) = a has a real solution, then the shortest interval containing 'a', is

If cos theta+sin theta=sqrt(2), then cos theta-sin theta=

If sin x cosh y = cos theta, cos x sinh y = sin theta then sinh ^ (2) y =

If the equation sin theta(sin theta + 2 cos theta )=a has a real solution then a belongs to the interval

If x=cos theta-cos2 theta,y=sin theta-sin2 theta then (dy)/(dx) is

Consider f(theta)=|[sin theta,cos theta,sin theta],[cos theta,sin theta,cos theta],[cos theta,sin theta,sin theta]| then A) f(theta)=0 has exactly 2 real solutions in [0,pi] B) f(theta)=0 has exactly 3 real solutions in [0,pi] C) Range of function (f(theta))/(1-sin2 theta) is [-sqrt(2),sqrt(2)] D) Range of function (f(theta))/(sin2 theta-1) is [-3,3]

If 2y cos theta =x sin theta and 2x sec theta -y cosec theta =3, then x ^(2) +4y^(2)=

if tan theta=(x)/(y) then sin2 theta+cos2 theta is