Let \theta , \phi \in \left[\right. 0 , 2 \pi \left]\right. be such that 2 cosθ \left(\right. 1 - sinϕ \left.\right) = \left(sin\right)^{2} \theta \left(\right. tan \frac{\theta}{2} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } cot \frac{\theta}{2} \left.\right) cosϕ - 1 , tan \left(\right. 2 \pi - \theta \left.\right) > 0 \textrm{ }\textrm{ } and \textrm{ }\textrm{ } - 1 < sinθ < \frac{- \sqrt{3}}{2} Then \phi cannot satisfy

Engineering

Mathematics

compound angle Formulae

Question

Let $θ, ϕ \in [0, 2 π]$ be such that

$2 cosθ (1 - sinϕ) = \sin^{2} θ (\tan \frac{θ}{2} + \cot \frac{θ}{2}) cosϕ - 1,$

$\tan (2 π - θ) > 0 and - 1 < sinθ < \frac{- \sqrt{3}}{2}$

Then $ϕ$ cannot satisfy

$\frac{3 π}{2} < ϕ < 2 π$

$\frac{4 π}{3} < ϕ < \frac{3 π}{2}$

$\frac{π}{2} < ϕ < \frac{4 π}{3}$

$0 < ϕ < \frac{π}{2}$

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$2 cosθ (1 - sinϕ) = \sin^{2} θ (\tan \frac{θ}{2} + \cot \frac{θ}{2}) cosϕ - 1$

tan $θ$ < 0 $\to θ$ lies in II and IV quadrant and – 1 < sin $θ < \frac{- \sqrt{3}}{2}$

$∴ θ$ lies in IV quadrant. $θ \in (\frac{3 π}{2}, \frac{5 π}{3})$

$2 cosθ (1 - sinϕ) = \sin^{2} θ (\frac{1}{\sin \frac{θ}{2} \cos \frac{θ}{2}}) cosϕ - 1$

$2 cosθ (1 - sinϕ) + 1 = \frac{4 \sin^{2} \frac{θ}{2} \cos^{2} \frac{θ}{2} \cdot cosϕ}{\sin \frac{θ}{2} \cos \frac{θ}{2}}$

2cos $θ$ (1 – sin $ϕ$ ) + 1 = 2sin $θ$ cos $ϕ$

2cos $θ$ + 1 = 2sin ( $θ + ϕ$ )

$θ \in (\frac{3 π}{2}, \frac{5 π}{3}) \Rightarrow cosθ \in (0, \frac{1}{2})$

cos $θ$ + 1 $\Rightarrow$ (1, 2)

2sin ( $θ + ϕ$ ) $\in$ (1, 2)

sin $(θ + ϕ) \in (\frac{1}{2}, 1)$

$(θ + ϕ) \in (\frac{13 π}{6}, \frac{5 π}{2}) or (\frac{5 π}{2}, \frac{17 π}{6})$

Since $θ \in (\frac{3 π}{2}, \frac{5 π}{3})$

By observation j does not belong to A, C and D.