If `piltalphalt(3pi)/2lt" then "sqrt((1-cosalpha)/(1+cosalpha))+sqrt((1+cosalpha)/(1-cosalpha))` is equal to

Options

A. `2/sinalpha`
B. `-2/sinalpha`
C. `1/sinalpha`
D. `-1/sinalpha`

Explanation

To solve the given expression \(\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}}\) under the condition that \(\alpha\) lies between \(\pi\) and \(\frac{3\pi}{2}\), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} \]

### Step 2: Find a common denominator
To combine the two square root terms, we can find a common denominator:
\[ \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = \frac{\sqrt{(1 - \cos \alpha)^2} + \sqrt{(1 + \cos \alpha)^2}}{\sqrt{(1 + \cos \alpha)(1 - \cos \alpha)}} \]

### Step 3: Simplify the numerator
The numerator simplifies to:
\[ \sqrt{(1 - \cos \alpha)^2} + \sqrt{(1 + \cos \alpha)^2} = |1 - \cos \alpha| + |1 + \cos \alpha| \]
Since \(\alpha\) is in the third quadrant, \(\cos \alpha\) is negative. Thus, \(1 - \cos \alpha\) is positive and \(1 + \cos \alpha\) is negative. Therefore:
\[ |1 - \cos \alpha| = 1 - \cos \alpha \quad \text{and} \quad |1 + \cos \alpha| = -(1 + \cos \alpha) \]
So, the numerator becomes:
\[ (1 - \cos \alpha) - (1 + \cos \alpha) = -2\cos \alpha \]

### Step 4: Simplify the denominator
The denominator simplifies as follows:
\[ \sqrt{(1 + \cos \alpha)(1 - \cos \alpha)} = \sqrt{1 - \cos^2 \alpha} = \sqrt{\sin^2 \alpha} = |\sin \alpha| \]
Since \(\alpha\) is in the third quadrant, \(\sin \alpha\) is negative, so:
\[ |\sin \alpha| = -\sin \alpha \]

### Step 5: Combine the results
Now we can combine the results from the numerator and denominator:
\[ \frac{-2\cos \alpha}{-\sin \alpha} = \frac{2\cos \alpha}{\sin \alpha} = 2 \cot \alpha \]

### Final Result
Thus, the expression simplifies to:
\[ \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = 2 \cot \alpha \]