Consider the function `f(x)=(sqrt(1+cos x)+sqrt(1-cosx))/(sqrt(1+cosx)-sqrt(1-cos x))` then
Q. If `x in (pi, 2pi)` then f(x) is

Options

A. `cot((pi)/(2)+(x)/(2))`
B. `tan((pi)/(4)+(x)/(2))`
C. `cot((pi)/(4)-(x)/(2))`
D. `tan(( pi)/(4)-(x)/(2))`

Answer

To solve the function \( f(x) = \frac{\sqrt{1 + \cos x} + \sqrt{1 - \cos x}}{\sqrt{1 + \cos x} - \sqrt{1 - \cos x}} \) for \( x \in (\pi, 2\pi) \), we can follow these steps:

### Step 1: Simplify the expression using trigonometric identities
We know that:
\[ \cos x = 1 - 2\sin^2\left(\frac{x}{2}\right) \]
and
\[ 1 - \cos x = 2\sin^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite \( \sqrt{1 + \cos x} \) and \( \sqrt{1 - \cos x} \):
\[ \sqrt{1 + \cos x} = \sqrt{1 + (1 - 2\sin^2\left(\frac{x}{2}\right))} = \sqrt{2 - 2\sin^2\left(\frac{x}{2}\right)} = \sqrt{2}\cos\left(\frac{x}{2}\right) \]
\[ \sqrt{1 - \cos x} = \sqrt{2\sin^2\left(\frac{x}{2}\right)} = \sqrt{2}\sin\left(\frac{x}{2}\right) \]

### Step 2: Substitute back into the function
Now substituting these back into \( f(x) \):
\[ f(x) = \frac{\sqrt{2}\cos\left(\frac{x}{2}\right) + \sqrt{2}\sin\left(\frac{x}{2}\right)}{\sqrt{2}\cos\left(\frac{x}{2}\right) - \sqrt{2}\sin\left(\frac{x}{2}\right)} \]
This simplifies to:
\[ f(x) = \frac{\cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)} \]

### Step 3: Factor out \(\sqrt{2}\)
Factoring out \(\sqrt{2}\) from both the numerator and denominator gives:
\[ f(x) = \frac{\cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)} \]

### Step 4: Analyze the function in the interval \( x \in (\pi, 2\pi) \)
In the interval \( x \in (\pi, 2\pi) \), we have \( \frac{x}{2} \in \left(\frac{\pi}{2}, \pi\right) \). In this range:
- \( \cos\left(\frac{x}{2}\right) \) is negative.
- \( \sin\left(\frac{x}{2}\right) \) is positive.

Thus, we can write:
\[ f(x) = \frac{\text{negative} + \text{positive}}{\text{negative} - \text{positive}} \]

### Step 5: Further simplify
This gives us:
\[ f(x) = \frac{\text{some positive value}}{\text{some negative value}} < 0 \]
Thus, \( f(x) \) is negative in the interval \( x \in (\pi, 2\pi) \).

### Conclusion
Therefore, the function \( f(x) \) is negative for \( x \in (\pi, 2\pi) \).