Let θ, φ ∈[0,2 π] be such that 2 cos θ(1- sin φ)= sin 2 θ( tan (θ/2)+ cot (θ/2)) cos φ-1, tan (2 π-θ)0 and -1

Question

Let θ, φ ∈[0,2 π] be such that 2 cos θ(1- sin φ)= sin 2 θ( tan (θ/2)+ cot (θ/2)) cos φ-1, tan (2 π-θ)>0 and -1< sin θ<-(√3/2). Then value of φ is (A) 0<φ<(π/2) (B) (π/2)<φ<(4 π/3) (C) (4 π/3)<φ<(3 π/2) (D) (3 π/2)<φ<2 π

Tardigrade Admin · Accepted Answer

text As tan (2 π-θ)>0,-1< sin θ<-(√3/2), θ ∈[0,2 π] ⇒ (3 π/2)<θ<(5 π/3) Now 2 cos θ(1- sin φ)= sin 2 θ( tan θ / 2+ cot θ / 2) cos φ-1 ⇒ 2 cos θ(1- sin φ)=2 sin θ cos φ-1 ⇒ 2 cos θ+1=2 sin (θ+φ) As θ ∈((3 π/2), (5 π/3)) ⇒ 2 cos θ+1 ∈(1,2) ⇒ 1<2 sin (θ+φ)< 2 ⇒ (1/2)< sin (θ+φ)<1 As θ+φ ∈[0,4 π] ⇒ θ+φ ∈((π/6), (5 π/6)) text or θ+φ ∈((13 π/6), (17 π/6)) ⇒ (π/6)-θ<φ<(5 π/6)-θ text or (13 π/6)-θ<φ<(17 π/6)-θ ⇒ φ ∈(-(3 π/2), (-2 π/3)) ∪((2 π/3), (7 π/6)) (∵ θ ∈((3 π/2), (5 π/3)))

Q. Let $θ, ϕ \in [0, 2 π]$ be such that $2 cos θ (1 - sin ϕ) = sin^{2} θ (tan \frac{θ}{2} + cot \frac{θ}{2}) cos ϕ - 1, tan (2 π - θ) > 0$ and $- 1 < sin θ < - \frac{3}{2}$ . Then value of $ϕ$ is

$0 < ϕ < \frac{π}{2}$

$\frac{π}{2} < ϕ < \frac{4 π}{3}$

$\frac{4 π}{3} < ϕ < \frac{3 π}{2}$

$\frac{3 π}{2} < ϕ < 2 π$

Q. Let θ,ϕ∈[0,2π] be such that 2cosθ(1−sinϕ)=sin2θ(tan2θ​+cot2θ​)cosϕ−1,tan(2π−θ)>0 and −1<sinθ<−23​​. Then value of ϕ is

0<ϕ<2π​

2π​<ϕ<34π​

34π​<ϕ<23π​

23π​<ϕ<2π

Q. Let $θ, ϕ \in [0, 2 π]$ be such that $2 cos θ (1 - sin ϕ) = sin^{2} θ (tan \frac{θ}{2} + cot \frac{θ}{2}) cos ϕ - 1, tan (2 π - θ) > 0$ and $- 1 < sin θ < - \frac{3}{2}$ . Then value of $ϕ$ is

$0 < ϕ < \frac{π}{2}$

$\frac{π}{2} < ϕ < \frac{4 π}{3}$

$\frac{4 π}{3} < ϕ < \frac{3 π}{2}$

$\frac{3 π}{2} < ϕ < 2 π$