Let `theta ,phi in [0,2pi]` be such that `2 cos theta (1-sin phi)=sin^2 theta ((tan)theta/2+(cot)theta/2) cos phi -1, tan(2pi-theta) > 0 and -1 < sin theta < -sqrt3/2.` Then `phi` cannot satisfy

Options

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

Answer

(a,c,d)

Explanation

PLAN It is based on range of sin x, i.e.
`[-1,1]` and the internal for `altxltb`.
Desription of Situtaion As `theta,phi in[0,2pi]` and
`tan(2pi-theta)gt0,-1sinthetalt-(sqrt(3))/(2)`
`tan(2pi-theta)gt0`
`implies" "-tanthetagt0`
`:." "theta in "II or IV quadrant."`
Also, `-1ltsintheta-(sqrt(3))/(2)`

`implies" "(4pi)/(3)ltthetalt(5pi)/(3)" but "theta in" II or IV quadrant"`
`implies" "(3pi)/(2)ltthetalt(5pi)/(3)" "...(i)`
Here, `2costheta(1-sinphi)=sin^(2)theta("tan"(theta)/(2)+"cot"(theta)/(2))cosphi-1`
`implies2costheta-2costhetasinphi=sin^(2)theta(("sin"^(2)(theta)/(2)+"cos"^(2)(theta)/(2))/("sin"(theta)/(2)"cos"(theta)/(2)))cosphi-1`
`implies2costheta-2costhetaphi=2sin^(2)theta((1)/(sintheta))cosphi-1`
`implies2costheta+1=2sinphicostheta+2sinthetacosphi`
`implies2costheta+1=2sin(theta+phi)" "...(ii)`
From Eq. (i), `(3pi)/(2)ltthetalt(5pi)/(3)`
`implies2costheta+1in(1,2)`
`:." "1lt2sin(theta+phi)lt2`
`implies(1)/(2)ltsin(theta+phi)lt1" "...(iii)`
`implies" "(pi)/(6)lttheta+philt(5pi)/(6)`
or `" "(13pi)/(6)lttheta+philt(17pi)/(6)`
`:." "(pi)/(6)-thetaltphilt(5pi)/(6)-theta`
or `" "(13pi)/(6)-thetaltphilt((17pi)/(6))-theta`
`impliesphi in(-(3pi)/(2),-(2pi)/(3))"or"((2pi)/(3),(7pi)/(6))," as "thetain((3pi)/(2),(5pi)/(3))`