If `pi le theta le (3pi)/2` , then `sqrt((1+cos theta)/(1-costheta))+sqrt((1-costheta)/(1+costheta))` is equal to :

Options

A. 2 cosec `theta`
B. `-2 "cosec" theta`
C. 2 sec `theta`
D. `-sec theta`

Answer

To solve the expression \( \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} + \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \) for \( \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2} \), we can follow these steps:

### Step 1: Combine the expressions
We start by combining the two square root terms over a common denominator:

\[ \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} + \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \frac{\sqrt{(1+\cos \theta)(1+\cos \theta)} + \sqrt{(1-\cos \theta)(1-\cos \theta)}}{\sqrt{(1-\cos \theta)(1+\cos \theta)}} \]

### Step 2: Simplify the numerator
The numerator simplifies to:

\[ \sqrt{(1+\cos \theta)^2} + \sqrt{(1-\cos \theta)^2} = |1+\cos \theta| + |1-\cos \theta| \]

### Step 3: Simplify the denominator
The denominator simplifies to:

\[ \sqrt{(1-\cos \theta)(1+\cos \theta)} = \sqrt{1 - \cos^2 \theta} = \sqrt{\sin^2 \theta} = |\sin \theta| \]

### Step 4: Analyze the absolute values
Since \( \theta \) is in the range \( \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2} \), we know:
- \( \cos \theta \) is negative.
- \( \sin \theta \) is negative in this range.

Thus, we have:
- \( |1+\cos \theta| = 1+\cos \theta \) (since \( \cos \theta < 0 \))
- \( |1-\cos \theta| = 1-\cos \theta \) (since \( 1 > \cos \theta \))

And for the denominator:
- \( |\sin \theta| = -\sin \theta \) (since \( \sin \theta < 0 \))

### Step 5: Substitute back into the expression
Now substituting these values back into our expression gives:

\[ \frac{(1+\cos \theta) + (1-\cos \theta)}{-\sin \theta} = \frac{2}{-\sin \theta} = -\frac{2}{\sin \theta} \]

### Step 6: Final result
Thus, we can express this as:

\[ -2 \csc \theta \]

### Conclusion
The final answer is:

\[ -2 \csc \theta \]