If `y = sqrt((cos x - sin x)(sec 2x + 1)(cos x + sin x))` ,
Then `(dy)/(dx)` at `x = (5 pi)/(6)` is :

Options

A. `(1)/(sqrt(2))`
B. `-(1)/(sqrt(2))`
C. `(sqrt(3))/(2)`
D. `-(sqrt(3))/(2)`

Explanation

To find \(\frac{dy}{dx}\) at \(x = \frac{5\pi}{6}\) for the function \(y = \sqrt{(\cos x - \sin x)(\sec 2x + 1)(\cos x + \sin x)}\), we will follow these steps:

### Step 1: Simplify the expression for \(y\)

We start with the expression:
\[ y = \sqrt{(\cos x - \sin x)(\sec 2x + 1)(\cos x + \sin x)} \]

Using the identity \(a^2 - b^2 = (a - b)(a + b)\), we can rewrite \((\cos x - \sin x)(\cos x + \sin x)\) as:
\[ \cos^2 x - \sin^2 x \]

Thus, we have:
\[ y = \sqrt{(\cos^2 x - \sin^2 x)(\sec 2x + 1)} \]

### Step 2: Substitute \(\sec 2x\)

Recall that \(\sec 2x = \frac{1}{\cos 2x}\). Therefore:
\[ \sec 2x + 1 = \frac{1 + \cos 2x}{\cos 2x} \]

Substituting this into our expression for \(y\):
\[ y = \sqrt{(\cos^2 x - \sin^2 x) \cdot \frac{1 + \cos 2x}{\cos 2x}} \]

### Step 3: Use the identity for \(1 + \cos 2x\)

Using the identity \(1 + \cos 2x = 2\cos^2 x\), we can substitute:
\[ y = \sqrt{(\cos^2 x - \sin^2 x) \cdot \frac{2 \cos^2 x}{\cos 2x}} \]

### Step 4: Simplify further

Using the identity \(\cos 2x = \cos^2 x - \sin^2 x\), we can simplify:
\[ y = \sqrt{(\cos^2 x - \sin^2 x) \cdot \frac{2 \cos^2 x}{\cos^2 x - \sin^2 x}} = \sqrt{2 \cos^2 x} \]

Thus, we have:
\[ y = \sqrt{2} \cdot |\cos x| \]

### Step 5: Evaluate \(y\) at \(x = \frac{5\pi}{6}\)

At \(x = \frac{5\pi}{6}\):
\[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]
Thus:
\[ y = \sqrt{2} \cdot \left| -\frac{\sqrt{3}}{2} \right| = \sqrt{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2} \]

### Step 6: Differentiate \(y\)

Now, we differentiate \(y\):
\[ y = \sqrt{2} \cdot |\cos x| \]

Since \(\cos x\) is negative in the second quadrant, we have:
\[ y = -\sqrt{2} \cdot \cos x \]

Differentiating:
\[ \frac{dy}{dx} = -\sqrt{2} \cdot (-\sin x) = \sqrt{2} \cdot \sin x \]

### Step 7: Evaluate \(\frac{dy}{dx}\) at \(x = \frac{5\pi}{6}\)

At \(x = \frac{5\pi}{6}\):
\[ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \]
Thus:
\[ \frac{dy}{dx} = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2} \]

### Final Answer

The value of \(\frac{dy}{dx}\) at \(x = \frac{5\pi}{6}\) is:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{2}} \]