Differentiate `tan^(-1){sqrt((1-cosx)/(1+cosx))} with respect to `x` :

Explanation

To differentiate the function \( y = \tan^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) \) with respect to \( x \), we can follow these steps:

### Step 1: Simplify the expression inside the arctangent
We start by simplifying the expression \( \sqrt{\frac{1 - \cos x}{1 + \cos x}} \).

Using the trigonometric identities:
- \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \)
- \( 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \)

We can rewrite the expression as:
\[ \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \sqrt{\frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)}} = \sqrt{\frac{\sin^2\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}} = \tan\left(\frac{x}{2}\right) \]

### Step 2: Rewrite \( y \)
Now we can rewrite \( y \):
\[ y = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) \]

### Step 3: Apply the property of inverse tangent
Since \( \tan^{-1}(\tan(\theta)) = \theta \) for \( \theta \) in the range of \( \tan^{-1} \), we have:
\[ y = \frac{x}{2} \]

### Step 4: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \]

### Final Answer
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \]

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