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. `thetaltphilt(pi)/(2)`
B. `(pi)/(2)ltphilt(4pi)/(3)`
C. `(4pi)/(3)lt philt(3pi)/(2)`
D. `(3pi)/(2)ltphilt2pi`

Answer

B

Explanation

We have,
`2costheta(1-sintheta)=sin^(2)theta(tantheta//2+cot theta//2)cos phi-1`
`implies2costheta(1-sinphi)=(2sin^(2)theta)/(sintheta)cosphi-1`
`implies2costheta+1=2sinthetacosphi+2costhetasinphi`
`implies2costheta+1=2sin(theta+phi)" "...(i)`
`Now, tan(2pi-theta)gt0and-1gtsinthetalt-(sqrt3)/(2)`
`implies-tanthetagt0and (4pi)/(3)ltthetalt(5pi)/(3)`
`impliesthetain((pi)/(2),pi)uu((3pi)/(2),2pi)and (4pi)/(3)ltthetalt(5pi)/(3)`
`implies(3pi)/(2)ltthetalt(5pi)/(3)`
From (i), we have
`sin(theta+phi)=costheta+1/2`
`implies1/2ltsin(theta+phi)lt1" "[{:(,because(3pi)/(2)ltthetalt(5pi)/(3)),(,0ltcos thetaimplies1/2lt1+costhetalt1):}]`
`implies2pi+(pi)/(6)lttheta+philt2pi+(5pi)/(6)`
`implies2pi+(pi)/(6)-theta_(max)ltphilt2pi+(5pi)/(6)-theta_(min)`
`implies2pi+(pi)/(6)-(5pi)/(3)ltphilt2pi+(5pi)/(6)-(3pi)/(2)implies(pi)/(2)ltphilt(4pi)/(3)`