solve the following expression `sqrt((1-costheta)/(1+costheta))+sqrt((1+costheta)/(1-costheta))` :

Options

A. `2sectheta`
B. `sectheta`
C. `2cosectheta`
D. none of these

Answer

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

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

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

### Step 3: Simplify the numerator
The numerator simplifies to:
\[ \sqrt{(1 - \cos \theta)^2} + \sqrt{(1 + \cos \theta)^2} = |1 - \cos \theta| + |1 + \cos \theta| \]
Since \( \cos \theta \) can range from -1 to 1, we analyze the cases:
1. If \( \cos \theta < 1 \), then \( |1 - \cos \theta| = 1 - \cos \theta \) and \( |1 + \cos \theta| = 1 + \cos \theta \).
2. If \( \cos \theta = 1 \), then both terms equal 0.

Thus, for \( \cos \theta < 1 \):
\[ |1 - \cos \theta| + |1 + \cos \theta| = (1 - \cos \theta) + (1 + \cos \theta) = 2 \]

### Step 4: Simplify the denominator
The denominator simplifies as follows:
\[ \sqrt{(1 + \cos \theta)(1 - \cos \theta)} = \sqrt{1 - \cos^2 \theta} = \sqrt{\sin^2 \theta} = |\sin \theta| \]
For \( \sin \theta \), we can assume \( \sin \theta \geq 0 \) in the context of this problem, so \( |\sin \theta| = \sin \theta \).

### Step 5: Combine the results
Now we can combine the results:
\[ \frac{2}{\sin \theta} \]

### Step 6: Final expression
Thus, the final expression simplifies to:
\[ 2 \cdot \frac{1}{\sin \theta} = 2 \csc \theta \]

### Conclusion
The simplified expression is:
\[ \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} + \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} = 2 \csc \theta \]