Let [ 0 , 2 ] be such that 2 cos ( 1 - sin ) = sin 2 ( tan2 + cot 2 ) cos - 1 , tan ( 2 ) 0 and - 1 sin - 3 2 , then cannot satisfy [2012]

Question

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 $- 1 < \sin θ < - \frac{\sqrt{3}}{2}$ , then $φ$ cannot satisfy [2012]

Smart Achievers · Accepted Answer

(1, 3, 4)

$As \tan (2 π - θ) > 0 and - 1 < \sin θ < - \frac{\sqrt{3}}{2}, θ \in [0, 2 π]$

$Hence \frac{3 π}{2} < θ < \frac{5 π}{3}$

$Now 2 \cos θ (1 - \sin φ) = \sin^{2} θ (\tan \frac{θ}{2} + \cot \frac{θ}{2}) \cos φ - 1$

$\Rightarrow 2 \cos θ (1 - \sin φ) = 2 \sin θ \cos φ - 1$

$\Rightarrow 2 \cos θ + 1 = 2 \sin (θ + φ)$

$As θ \in (\frac{3 π}{2}, \frac{5 π}{3}), 1 < 2 \sin (θ + φ) < 2$

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

$We have φ \in (- \frac{3 π}{2}, - \frac{2 π}{3}) \cup (\frac{2 π}{3}, \frac{7 π}{6})$

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