Let `varphi,phi in [0,2pi]` be such that `2costheta(1-sinphi)=sin^2theta(tantheta/2+cottheta//2)"cos"phi-1,"tan"(2pi-theta)>0` `and −1<"sinθ"< −sqrt3/2 ` then φ lies between

Options

A. `0 lt phi lt pi/2`
B. `pi/2 lt phi lt (4pi)/3`
C. `(4pi)/3 lt phi lt (3pi)/2`
D. `(3pi)/2 lt phi lt 2pi`

Answer

A, C, D

Explanation

`2 cos theta (1- sin phi)=(2 sin^(2) theta)/( sin theta) cos phi -1`
`=2 sin cos phi -1`
`:. 2 cos theta-2 cos theta sin phi=2 sin theta cos phi-1`
`:. 2 cos theta +1 =2 sin (theta+phi)` ...(i)
`tan (2pi -theta) gt 0`
`rArr tan theta lt 0`
and `-1 lt sin theta lt - sqrt(3)/2`
`rArr theta in ((3pi)/2, (5pi)/3)`
`rArr 0 lt cos theta lt 1/2`
`rArr 1/2 lt sin (theta + phi) lt 1` (from (i))
`rArr pi/6+2pi lt theta + phi lt (5pi)/6 +2pi`
`2pi +pi/6- theta_("max") lt phi lt 2pi + (5pi)/6 - theta_(min)`
`rArr pi/2 lt phi lt (4pi)/3`