If `f (x) = tan ^(-1)sqrt((1 + sin x )/(1 - sin x)), 0 le x le (pi)/(2) `then `f' ((pi)/(6))` =?

Options

A. `-(1)/(4)`
B. `-(1)/(2)`
C. `(1)/(4)`
D. `(1)/(2)`

Answer

To solve the problem, we need to find the derivative \( f'(\frac{\pi}{6}) \) where

\[ f(x) = \tan^{-1} \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right) \]

### Step 1: Simplify the function \( f(x) \)

We start with the expression inside the arctangent:

\[ \sqrt{\frac{1 + \sin x}{1 - \sin x}} \]

Using the identity \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \), we can rewrite \( 1 + \sin x \) and \( 1 - \sin x \):

\[ 1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)^2 \]

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

Thus, we can rewrite \( f(x) \) as:

\[ f(x) = \tan^{-1} \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right) \]

### Step 2: Recognize the tangent function

Notice that:

\[ \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \]

Thus, we can express \( f(x) \) as:

\[ f(x) = \tan^{-1} \left( \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right) = \frac{\pi}{4} + \frac{x}{2} \]

### Step 3: Differentiate \( f(x) \)

Now we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \frac{\pi}{4} + \frac{x}{2} \right) = \frac{1}{2} \]

### Step 4: Evaluate \( f'(\frac{\pi}{6}) \)

Since \( f'(x) = \frac{1}{2} \) is constant, we have:

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

### Final Answer

Thus, the value of \( f' \left( \frac{\pi}{6} \right) \) is:

\[ \boxed{\frac{1}{2}} \]