Consider `f(x)=tan^(-1)(sqrt((1+sinx)/(1-sinx))), x in (0,pi/2)dot` A normal to `y=f(x)` at `x=pi/6` also passes through the point: (1) (0, 0) (2) `(0,(2pi)/3)` (3) `(pi/6,0)` (4) `(pi/4,0)`

Options

A. `(0,(2pi)/(3))`
B. `((pi)/(6)0,)`
C. `((pi)/(4),0)`
D. `(0,0)`

Answer

To solve the problem, we need to find the equation of the normal to the curve \( y = f(x) \) at \( x = \frac{\pi}{6} \) and determine which of the given points lies on this normal line.

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

### Step 2: Find \( f\left(\frac{\pi}{6}\right) \)
First, we need to evaluate \( f\left(\frac{\pi}{6}\right) \):
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
So,
\[ f\left(\frac{\pi}{6}\right) = \tan^{-1}\left(\sqrt{\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}}\right) = \tan^{-1}\left(\sqrt{\frac{\frac{3}{2}}{\frac{1}{2}}}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Thus, the point on the curve at \( x = \frac{\pi}{6} \) is \( \left(\frac{\pi}{6}, \frac{\pi}{3}\right) \).

### Step 3: Find \( f'(x) \)
Next, we differentiate \( f(x) \):
Using the chain rule:
\[ f'(x) = \frac{1}{1 + \left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)^2} \cdot \frac{d}{dx}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right) \]
Calculating the derivative of the inside function using the quotient rule:
Let \( u = 1 + \sin x \) and \( v = 1 - \sin x \):
\[ \frac{d}{dx}\left(\sqrt{\frac{u}{v}}\right) = \frac{1}{2\sqrt{\frac{u}{v}}} \cdot \frac{v \cdot u' - u \cdot v'}{v^2} \]
Where \( u' = \cos x \) and \( v' = -\cos x \):
\[ = \frac{1}{2\sqrt{\frac{1 + \sin x}{1 - \sin x}}} \cdot \frac{(1 - \sin x) \cos x + (1 + \sin x) \cos x}{(1 - \sin x)^2} \]
After simplification, we find:
\[ f'(x) = \frac{1}{2(1 + \sin x)} \]

### Step 4: Evaluate \( f'\left(\frac{\pi}{6}\right) \)
Now we evaluate the derivative at \( x = \frac{\pi}{6} \):
\[ f'\left(\frac{\pi}{6}\right) = \frac{1}{2(1 + \frac{1}{2})} = \frac{1}{2 \cdot \frac{3}{2}} = \frac{1}{3} \]

### Step 5: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the derivative:
\[ \text{slope of normal} = -\frac{1}{f'\left(\frac{\pi}{6}\right)} = -3 \]

### Step 6: Write the equation of the normal line
Using the point-slope form of the line:
\[ y - \frac{\pi}{3} = -3\left(x - \frac{\pi}{6}\right) \]
Simplifying gives:
\[ y = -3x + \frac{3\pi}{6} + \frac{\pi}{3} = -3x + \frac{2\pi}{3} \]
Rearranging:
\[ 3x + y = \frac{2\pi}{3} \]

### Step 7: Check which points lie on the normal line
Now we check the given points:
1. For \( (0, 0) \):
\[ 3(0) + 0 = 0 \quad \text{(not equal to } \frac{2\pi}{3}\text{)} \]
2. For \( \left(0, \frac{2\pi}{3}\right) \):
\[ 3(0) + \frac{2\pi}{3} = \frac{2\pi}{3} \quad \text{(this point lies on the normal)} \]
3. For \( \left(\frac{\pi}{6}, 0\right) \):
\[ 3\left(\frac{\pi}{6}\right) + 0 = \frac{\pi}{2} \quad \text{(not equal to } \frac{2\pi}{3}\text{)} \]
4. For \( \left(\frac{\pi}{4}, 0\right) \):
\[ 3\left(\frac{\pi}{4}\right) + 0 = \frac{3\pi}{4} \quad \text{(not equal to } \frac{2\pi}{3}\text{)} \]

### Conclusion
The only point that lies on the normal line is \( (0, \frac{2\pi}{3}) \).