If `2y=(cot^(-1)((sqrt3 cos x + sin x)/(cos x - sqrt3 sin x)))^(2) , x in (0, pi/2) " then " (dy)/(dx) ` is equal to

Options

A. `( pi)/( 6) -x `
B. `2x - ( pi)/( 3)`
C. ` x - ( pi)/( 6)`
D. `(pi)/( 3) - x `

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given the equation:

\[ 2y = \left( \cot^{-1} \left( \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) \right)^2 \]

### Step 1: Simplify the expression inside the cotangent inverse

We start with the expression inside the cotangent inverse:

\[ \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \]

To simplify this, we can divide the numerator and the denominator by \(\cos x\):

\[ = \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x} \]

### Step 2: Recognize the angle in terms of tangent

The expression \(\frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x}\) can be recognized as the tangent of a sum of angles. Specifically:

\[ \tan\left(\frac{\pi}{3} + x\right) = \frac{\tan\frac{\pi}{3} + \tan x}{1 - \tan\frac{\pi}{3} \tan x} = \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x} \]

Thus, we have:

\[ \cot^{-1}\left(\frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x}\right) = \frac{\pi}{3} + x \]

### Step 3: Substitute back into the equation

Now substituting back into our equation for \(2y\):

\[ 2y = \left(\frac{\pi}{3} + x\right)^2 \]

### Step 4: Differentiate both sides with respect to \(x\)

Now we differentiate both sides with respect to \(x\):

\[ \frac{d}{dx}(2y) = \frac{d}{dx}\left(\left(\frac{\pi}{3} + x\right)^2\right) \]

Using the chain rule on the left side:

\[ 2 \frac{dy}{dx} = 2\left(\frac{\pi}{3} + x\right) \cdot \frac{d}{dx}\left(\frac{\pi}{3} + x\right) \]

Since \(\frac{d}{dx}\left(\frac{\pi}{3} + x\right) = 1\):

\[ 2 \frac{dy}{dx} = 2\left(\frac{\pi}{3} + x\right) \]

### Step 5: Solve for \(\frac{dy}{dx}\)

Dividing both sides by 2 gives:

\[ \frac{dy}{dx} = \frac{\pi}{3} + x \]

### Final Answer

Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{\pi}{3} + x \]