Let θ, ϕ ∈ [0, 2] be such that 2cosθ(1 - sinϕ) = sin2θ(tan2θ(tan(θ/2) + cot(θ/2))cosϕ - 1, tan(2π - θ) > 0 and - 1 < sinθ < - √3/2. Then ϕ cannot satisfy

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`