If `f (x) = sqrt(cos ec ^(2) x - 2 sin x cos x - (1)/(tan ^(2) x ))`
`x in ((7pi)/(4), 2pi )` then ` f' ((11 pi)/(6))=`

Options

A. `(1- sqrt3)/(2)`
B. `(sqrt3-1)/(2)`
C. `(1+sqrt3)/(2)`
D. None of these

Explanation

To solve the problem step by step, we will analyze the function \( f(x) \) and then find its derivative at the specified point.

### Step 1: Rewrite the function
Given:
\[ f(x) = \sqrt{\cos^2 x - 2 \sin x \cos x - \frac{1}{\tan^2 x}} \]
We can rewrite \( \frac{1}{\tan^2 x} \) as \( \cot^2 x \):
\[ f(x) = \sqrt{\cos^2 x - 2 \sin x \cos x - \cot^2 x} \]

### Step 2: Simplify the expression
Using the identity \( \cot^2 x = \frac{\cos^2 x}{\sin^2 x} \), we can rewrite \( f(x) \):
\[ f(x) = \sqrt{\cos^2 x - 2 \sin x \cos x - \frac{\cos^2 x}{\sin^2 x}} \]
Now, we can combine the terms:
\[ f(x) = \sqrt{\cos^2 x - 2 \sin x \cos x - \frac{\cos^2 x}{\sin^2 x}} \]

### Step 3: Substitute \( \sin^2 x + \cos^2 x = 1 \)
We know that \( \sin^2 x + \cos^2 x = 1 \). Therefore, we can express \( 1 \) as \( \sin^2 x + \cos^2 x \):
\[ f(x) = \sqrt{(\cos^2 x + \sin^2 x) - 2 \sin x \cos x} \]
This simplifies to:
\[ f(x) = \sqrt{1 - 2 \sin x \cos x} \]

### Step 4: Use the double angle identity
Using the double angle identity \( \sin(2x) = 2 \sin x \cos x \):
\[ f(x) = \sqrt{1 - \sin(2x)} \]

### Step 5: Find the derivative \( f'(x) \)
To find \( f'(x) \), we apply the chain rule:
\[ f'(x) = \frac{1}{2\sqrt{1 - \sin(2x)}} \cdot (-\cos(2x) \cdot 2) \]
Thus,
\[ f'(x) = -\frac{\cos(2x)}{\sqrt{1 - \sin(2x)}} \]

### Step 6: Evaluate \( f' \left( \frac{11\pi}{6} \right) \)
Now we need to evaluate \( f' \left( \frac{11\pi}{6} \right) \):
1. Calculate \( \sin(2 \cdot \frac{11\pi}{6}) = \sin\left(\frac{11\pi}{3}\right) = \sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2} \)
2. Calculate \( \cos(2 \cdot \frac{11\pi}{6}) = \cos\left(\frac{11\pi}{3}\right) = \cos\left(\frac{5\pi}{3}\right) = \frac{1}{2} \)

Now substituting these values into the derivative:
\[ f' \left( \frac{11\pi}{6} \right) = -\frac{\frac{1}{2}}{\sqrt{1 - (-\frac{\sqrt{3}}{2})}} = -\frac{\frac{1}{2}}{\sqrt{1 + \frac{\sqrt{3}}{2}}} \]

### Step 7: Simplify the expression
To simplify \( \sqrt{1 + \frac{\sqrt{3}}{2}} \):
\[ 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \]
Thus,
\[ \sqrt{1 + \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{2}} = \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2}} \]
So,
\[ f' \left( \frac{11\pi}{6} \right) = -\frac{\frac{1}{2}}{\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2}}} = -\frac{\sqrt{2}}{2\sqrt{2 + \sqrt{3}}} \]

### Final Answer
Thus, the value of \( f' \left( \frac{11\pi}{6} \right) \) is:
\[ f' \left( \frac{11\pi}{6} \right) = 1 - \frac{\sqrt{3}}{2} \]