If `f(x) = tan^(-1)(sqrt((1+sinx)/(1-sinx))), 0 lt x lt pi/2`, then `f'(pi/6)` is

Options

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

Answer

To find \( f'(\frac{\pi}{6}) \) for the function \( f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right) \), we will follow these steps:

### Step 1: Simplify the function
We start with the function:
\[ f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right) \]

Using the identity:
\[ \frac{1+\sin x}{1-\sin x} = \frac{\cos^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} = \cot^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite:
\[ f(x) = \tan^{-1}\left(\sqrt{\cot^2\left(\frac{x}{2}\right)}\right) = \tan^{-1}\left(\cot\left(\frac{x}{2}\right)\right) \]

Since \( \tan^{-1}(\cot(\theta)) = \frac{\pi}{2} - \theta \), we have:
\[ f(x) = \frac{\pi}{2} - \frac{x}{2} \]

### Step 2: Differentiate the function
Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}\left(\frac{\pi}{2} - \frac{x}{2}\right) = 0 - \frac{1}{2} = -\frac{1}{2} \]

### Step 3: Evaluate the derivative at \( x = \frac{\pi}{6} \)
Now, we find \( f'(\frac{\pi}{6}) \):
\[ f'(\frac{\pi}{6}) = -\frac{1}{2} \]

### Final Answer
Thus, the value of \( f'(\frac{\pi}{6}) \) is:
\[ \boxed{-\frac{1}{2}} \]