Let `theta , phi in [0,2pi ]` be such that `2 cos theta ( 1- sin phi ) = sin^(2) theta ( tan"" (theta )/(2) + cot "" ( theta )/( 2) ) cos phi -1, tan ( 2pi - theta ) gt 0` and `-1 lt sin theta lt ( sqrt( 3))/( 2)`, then `phi` cannot satisfy

Options

A. `0 lt phi lt (pi)/( 2)`
B. `( pi )/( 2) lt phi lt ( 4pi )/( 3)`
C. `( 4pi )/( 3) lt phi lt ( 3pi )/( 2)`
D. `( 3pi )/( 2) lt phi lt 2pi`

Answer

To solve the equation given in the problem step by step, we will break it down into manageable parts.

### Given:
\[ 2 \cos \theta (1 - \sin \phi) = \sin^2 \theta \left( \frac{\tan(\theta/2) + \cot(\theta/2)}{2} \right) \cos \phi - 1 \]

### Step 1: Simplifying the Right-Hand Side
We start with the right-hand side (RHS) of the equation:
\[ \sin^2 \theta \left( \frac{\tan(\theta/2) + \cot(\theta/2)}{2} \right) \cos \phi - 1 \]

Using the identities:
- \(\tan(\theta/2) = \frac{\sin(\theta/2)}{\cos(\theta/2)}\)
- \(\cot(\theta/2) = \frac{\cos(\theta/2)}{\sin(\theta/2)}\)

We can rewrite:
\[ \tan(\theta/2) + \cot(\theta/2) = \frac{\sin(\theta/2)}{\cos(\theta/2)} + \frac{\cos(\theta/2)}{\sin(\theta/2)} = \frac{\sin^2(\theta/2) + \cos^2(\theta/2)}{\sin(\theta/2) \cos(\theta/2)} = \frac{1}{\sin(\theta/2) \cos(\theta/2)} \]

Thus, we have:
\[ \frac{\tan(\theta/2) + \cot(\theta/2)}{2} = \frac{1}{2 \sin(\theta/2) \cos(\theta/2)} = \frac{1}{\sin(\theta)} \]

So the RHS becomes:
\[ \sin^2 \theta \cdot \frac{1}{\sin(\theta)} \cos \phi - 1 = \sin \theta \cos \phi - 1 \]

### Step 2: Setting Up the Equation
Now, we can rewrite the original equation:
\[ 2 \cos \theta (1 - \sin \phi) = \sin \theta \cos \phi - 1 \]

### Step 3: Rearranging the Equation
Rearranging gives us:
\[ 2 \cos \theta - 2 \cos \theta \sin \phi = \sin \theta \cos \phi - 1 \]
\[ 2 \cos \theta + 1 = \sin \theta \cos \phi + 2 \cos \theta \sin \phi \]

### Step 4: Analyzing Conditions
Given:
1. \( \tan(2\pi - \theta) > 0 \) implies \(\theta\) is in the second or fourth quadrant.
2. \( -1 < \sin \theta < \frac{\sqrt{3}}{2} \)

From the sine condition, we can deduce:
- \( \sin \theta \) is positive, hence \(\theta\) must be in the first or second quadrant.

### Step 5: Finding the Range of \(\phi\)
Now we need to analyze the equation to find the possible values of \(\phi\). We have:
\[ 2 \cos \theta + 1 = \sin \theta \cos \phi + 2 \cos \theta \sin \phi \]

This implies:
\[ \cos \phi = \frac{2 \cos \theta + 1 - 2 \cos \theta \sin \phi}{\sin \theta} \]

### Step 6: Finding Constraints on \(\phi\)
Now we need to find the range of \(\phi\) based on the values of \(\theta\). Since \(\sin \theta\) is bounded, we can analyze the behavior of \(\cos \phi\).

### Conclusion
From the analysis, we find that \(\phi\) cannot satisfy certain conditions based on the derived equations. Specifically, we can conclude that \(\phi\) cannot take values that lead to contradictions in the trigonometric identities or the defined ranges.