Question

Let $\theta ,\varphi \in [0,2\pi ]$ be such that $2cos\theta (1-sin\varphi )=si{n}^2\theta (tan\frac{\theta }{2}+cot\frac{\theta }{2})cos\theta -1,tan(2\pi -\theta )>0$ and -1 < $sin\theta <-\frac{\sqrt{3}}{2}$. Then $\varphi$ cannot satisfy

• A. $0<\varphi <\frac{\pi }{2}$
  
• B. $\frac{\pi }{2}<\varphi <\frac{4\pi }{3}$
  
• C. $\frac{4\pi }{3}<\varphi <\frac{3\pi }{2}$
  
• D. $\frac{3\pi }{2}<\varphi <2\pi$

Answer 1

Answer: (A), (C), (D)

$\frac{3\pi }{2}<\varphi <2\pi$

$2cos\theta (1-sin\varphi )=\frac{2si{n}^2\theta }{sin\theta }cos\varphi -1\therefore 2cos\theta -2cos\theta sin\varphi =2sin\theta cos\varphi -1\therefore 2cos\theta +1=2sin(\theta +\varphi )tan(2\pi -\theta )>0\Rightarrow tan\theta <0$
and $-1<sin\theta <-\frac{\sqrt{3}}{2}$
$\Rightarrow \theta \in (\frac{3\pi }{2},\frac{5\pi }{3})$
$\Rightarrow 0<cos\theta <\frac{1}{2}$
$\frac{1}{2}<sin(\theta +\varphi )<1$
$\Rightarrow \frac{\pi }{6}+2\pi <sin(\theta +\varphi )<\frac{5\pi }{6}+2\pi$
$2\pi +\frac{\pi }{6}-{\theta }_{max}<\varphi <2\pi +\frac{5\pi }{6}-{\theta }_{min}$
$\Rightarrow \frac{\pi }{2}<\varphi \frac{4\pi }{3}$