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

(A) $0<\varphi<\frac{\pi}{2}$; (C) $\frac{4 \pi}{3}<\varphi<\frac{3 \pi}{2}$; (D) $\frac{3 \pi}{2}<\varphi<2 \pi$

(B) $\frac{\pi}{2}<\varphi<\frac{4 \pi}{3}$

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Trigonometric Equations

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

φ

cannot satisfy

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

Simplifying the given equation:

tan \frac{θ}{2} + cot \frac{θ}{2} = \frac{2}{s i n θ}

, so

sin^{2} θ \cdot \frac{2}{s i n θ} cos φ = 2 sin θ cos φ

. The equation becomes

2 cos θ - 2 cos θ sin φ = 2 sin θ cos φ - 1

, which simplifies to

2 cos θ + 1 = 2 sin (θ + φ)

. From

tan (2 π - θ) > 0

, we have

tan θ < 0

. Combined with

- 1 < sin θ < - \frac{3}{2}

, we get

θ \in (\frac{3 π}{2}, \frac{5 π}{3})

. Then

\frac{1}{2} < sin (θ + φ) < 1

, giving

\frac{π}{6} < θ + φ < \frac{5 π}{6}

(mod

2 π

). This leads to

\frac{π}{2} < φ < \frac{4 π}{3}

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · One or more correct options