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 find \( f' \left( \frac{\pi}{6} \right) \) for the function

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

we will first simplify the function \( f(x) \).

### Step 1: Simplifying \( f(x) \)

Using the identity for tangent of half-angle, we know that:

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

We can rewrite \( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \) using the half-angle identities.

Using the identity:

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

and

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

we can simplify \( f(x) \):

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

This is equivalent to:

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

### Step 2: Finding the derivative \( f'(x) \)

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

\[ f'(x) = \frac{d}{dx} \left( \frac{\pi}{4} + \frac{x}{2} \right) = 0 + \frac{1}{2} = \frac{1}{2} \]

### Step 3: Evaluating \( f' \left( \frac{\pi}{6} \right) \)

Since \( f'(x) \) 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

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