If `f(x)=sqrt("cosec"^2x-2sinx cosx - 1/(tan^2x)) x in ((7pi)/4,2pi)`, then `f '((11pi)/6)=`

Options

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

Explanation

To solve the problem, we need to find \( f'(\frac{11\pi}{6}) \) for the function defined as:

\[ f(x) = \sqrt{\csc^2 x - 2 \sin x \cos x - \frac{1}{\tan^2 x}} \]

### Step 1: Simplify the expression inside the square root

We start with the expression:

\[ f(x) = \sqrt{\csc^2 x - 2 \sin x \cos x - \frac{1}{\tan^2 x}} \]

Recall that:

\[ \csc^2 x = \frac{1}{\sin^2 x} \quad \text{and} \quad \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \implies \frac{1}{\tan^2 x} = \frac{\cos^2 x}{\sin^2 x} \]

Substituting these into the function gives:

\[ f(x) = \sqrt{\frac{1}{\sin^2 x} - 2 \sin x \cos x - \frac{\cos^2 x}{\sin^2 x}} \]

Combine the terms:

\[ f(x) = \sqrt{\frac{1 - 2 \sin x \cos x \sin^2 x - \cos^2 x}{\sin^2 x}} \]

### Step 2: Use trigonometric identities

Using the identity \( \sin^2 x + \cos^2 x = 1 \):

\[ f(x) = \sqrt{\frac{1 - \cos^2 x - 2 \sin x \cos x}{\sin^2 x}} = \sqrt{\frac{\sin^2 x - 2 \sin x \cos x}{\sin^2 x}} \]

This simplifies to:

\[ f(x) = \sqrt{\frac{(\sin x - \cos x)^2}{\sin^2 x}} = \frac{|\sin x - \cos x|}{|\sin x|} \]

### Step 3: Determine the sign in the interval

Since \( x \) is in the interval \( \left(\frac{7\pi}{4}, 2\pi\right) \), we know:

- \( \sin x < 0 \)
- \( \cos x > 0 \)

Thus, \( \sin x - \cos x < 0 \) in this interval, so:

\[ f(x) = -(\sin x - \cos x) = \cos x - \sin x \]

### Step 4: Differentiate \( f(x) \)

Now we differentiate:

\[ f'(x) = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x \]

### Step 5: Evaluate \( f'(\frac{11\pi}{6}) \)

Now we need to evaluate this derivative at \( x = \frac{11\pi}{6} \):

\[ f'\left(\frac{11\pi}{6}\right) = -\sin\left(\frac{11\pi}{6}\right) - \cos\left(\frac{11\pi}{6}\right) \]

Using the unit circle values:

\[ \sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}, \quad \cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

Substituting these values gives:

\[ f'\left(\frac{11\pi}{6}\right) = -\left(-\frac{1}{2}\right) - \frac{\sqrt{3}}{2} = \frac{1}{2} - \frac{\sqrt{3}}{2} \]

### Final Answer

Thus, the final answer is:

\[ f'\left(\frac{11\pi}{6}\right) = \frac{1 - \sqrt{3}}{2} \]