Let θ, ∅ ∈[0,2π] be such that 2cos θ 1-sin ∅ = sin2 θ tanθ2+ cotθ2 cos ∅-1,tan(2π - θ) >0 and -1< sin θ <-32 . then ∅ can’t satisfy

tan(2π−θ)>0⇒−tanθ>0⇒tanθ<0 ⇒θ∈II or IV quadrant−1<sinθ<−32⇒4π3<θ<5π3−{3π2} ⇒3π2<θ<5π3.......(i)2cosθ(1−sinϕ)=sin2θtanθ2+cotθ2cosϕ−1 ⇒2cosθ−2cosθ.sinϕ=sin2θ(sin2θ2+cos2θ/2sinθ/2.cosθ/2)cosϕ−1 ⇒2cosθ−2cosθ.sinϕ=2sin2θ(1sinθ).cosϕ−1⇒2cosθ+1=2sin(θ+ϕ) 3π2<θ<5π3⇒2cosθ+1∈(1,2)1<2sin(θ+ϕ)<2⇒12<sin(θ+ϕ)<1 ⇒π6<θ+ϕ<5π6 or 13π6<θ+ϕ<17π6−{π2}⇒π6−θ<ϕ<5π6−θ or 13π6−θ<ϕ<17π6−θ⇒ϕ∈(−3π2,−2π3) or (2π3,7π6)

Let θ, ∅ ∈[0,2π] be such that 2cos θ 1-sin ∅ = sin2 θ tanθ2+ cotθ2 cos ∅-1,tan(2π - θ) >0 and -1< sin θ <-32 . then ∅ can’t satisfy

tan(2π−θ)>0⇒−tanθ>0⇒tanθ<0 ⇒θ∈II or IV quadrant−1<sinθ<−32⇒4π3<θ<5π3−{3π2} ⇒3π2<θ<5π3.......(i)2cosθ(1−sinϕ)=sin2θtanθ2+cotθ2cosϕ−1 ⇒2cosθ−2cosθ.sinϕ=sin2θ(sin2θ2+cos2θ/2sinθ/2.cosθ/2)cosϕ−1 ⇒2cosθ−2cosθ.sinϕ=2sin2θ(1sinθ).cosϕ−1⇒2cosθ+1=2sin(θ+ϕ) 3π2<θ<5π3⇒2cosθ+1∈(1,2)1<2sin(θ+ϕ)<2⇒12<sin(θ+ϕ)<1 ⇒π6<θ+ϕ<5π6 or 13π6<θ+ϕ<17π6−{π2}⇒π6−θ<ϕ<5π6−θ or 13π6−θ<ϕ<17π6−θ⇒ϕ∈(−3π2,−2π3) or (2π3,7π6)

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Let $θ, \emptyset \in [0, 2 π]$ be such that $2 \cos θ (1 - \sin \emptyset) = \sin^{2} θ (\tan \frac{θ}{2} + c o t \frac{θ}{2}) \cos \emptyset - 1$ ,
$\tan (2 π - θ) > 0 a n d - 1 < \sin θ < - \frac{\sqrt{3}}{2}$ . then $\emptyset$ can’t satisfy

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$\frac{3 π}{2} < \emptyset < 2 π$

$0 < \emptyset < \frac{π}{2}$

$\frac{π}{2} < \emptyset < \frac{4 π}{3}$

$\frac{4 π}{3} < \emptyset < \frac{3 π}{2}$

answer is A, C, D.

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Detailed Solution

$\tan (2 π - θ) > 0 \Rightarrow - \tan θ > 0 \Rightarrow \tan θ < 0 \begin{array}{l} \Rightarrow θ \in I I o r I V q u a d r a n t \\ - 1 < \sin θ < - \frac{\sqrt{3}}{2} \Rightarrow \frac{4 π}{3} < θ < \frac{5 π}{3} - {\frac{3 π}{2}} \end{array} \begin{array}{l} \Rightarrow \frac{3 π}{2} < θ < \frac{5 π}{3} ....... (i) \\ 2 \cos θ (1 - \sin ϕ) = \sin^{2} θ (\tan \frac{θ}{2} + \cot \frac{θ}{2}) \cos ϕ - 1 \end{array} \Rightarrow 2 \cos θ - 2 \cos θ . \sin ϕ = \sin^{2} θ (\frac{\sin^{2} \frac{θ}{2} + \cos^{2} θ / 2}{\sin θ / 2. \cos θ / 2}) \cos ϕ - 1 \begin{array}{l} \Rightarrow 2 \cos θ - 2 \cos θ . \sin ϕ = 2 \sin^{2} θ (\frac{1}{\sin θ}) . \cos ϕ - 1 \\ \Rightarrow 2 \cos θ + 1 = 2 \sin (θ + ϕ) \end{array} \begin{array}{l} \frac{3 π}{2} < θ < \frac{5 π}{3} \Rightarrow 2 \cos θ + 1 \in (1, 2) \\ 1 < 2 \sin (θ + ϕ) < 2 \Rightarrow \frac{1}{2} < \sin (θ + ϕ) < 1 \end{array} \begin{array}{l} \Rightarrow \frac{π}{6} < θ + ϕ < \frac{5 π}{6} o r \frac{13 π}{6} < θ + ϕ < \frac{17 π}{6} - {\frac{π}{2}} \\ \Rightarrow \frac{π}{6} - θ < ϕ < \frac{5 π}{6} - θ o r \frac{13 π}{6} - θ < ϕ < \frac{17 π}{6} - θ \\ \Rightarrow ϕ \in (- \frac{3 π}{2}, - \frac{2 π}{3}) o r (\frac{2 π}{3}, \frac{7 π}{6}) \end{array}$