Find the maximum and minimum values of the function `f(x) = (sin x)/(1+ tan x) ,(0 lt x lt 2pi)`.

Answer

To find the maximum and minimum values of the function \( f(x) = \frac{\sin x}{1 + \tan x} \) in the interval \( (0, 2\pi) \), we will follow these steps:

### Step 1: Differentiate the function
We will use the quotient rule for differentiation. The quotient rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

For our function, let:
- \( u = \sin x \) and \( u' = \cos x \)
- \( v = 1 + \tan x \) and \( v' = \sec^2 x \)

Thus, we have:

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

### Step 2: Set the derivative equal to zero
To find the critical points, we set the numerator equal to zero:

\[ \cos x (1 + \tan x) - \sin x \sec^2 x = 0 \]

This simplifies to:

\[ \cos x + \sin x - \sin x = 0 \]

So we have:

\[ \cos x = 0 \]

### Step 3: Solve for critical points
The solutions to \( \cos x = 0 \) in the interval \( (0, 2\pi) \) are:

\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \]

### Step 4: Evaluate the function at critical points and endpoints
Next, we evaluate \( f(x) \) at the critical points and also check the endpoints of the interval \( (0, 2\pi) \):

1. **At \( x = \frac{\pi}{2} \)**:
\[ f\left(\frac{\pi}{2}\right) = \frac{\sin\left(\frac{\pi}{2}\right)}{1 + \tan\left(\frac{\pi}{2}\right)} = \frac{1}{1 + \infty} = 0 \]

2. **At \( x = \frac{3\pi}{2} \)**:
\[ f\left(\frac{3\pi}{2}\right) = \frac{\sin\left(\frac{3\pi}{2}\right)}{1 + \tan\left(\frac{3\pi}{2}\right)} = \frac{-1}{1 + \infty} = 0 \]

3. **At \( x = 0 \)**:
\[ f(0) = \frac{\sin(0)}{1 + \tan(0)} = \frac{0}{1} = 0 \]

4. **At \( x = 2\pi \)**:
\[ f(2\pi) = \frac{\sin(2\pi)}{1 + \tan(2\pi)} = \frac{0}{1} = 0 \]

### Step 5: Check values in between the critical points
To find the maximum and minimum values, we can also check values between the critical points, such as \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \):

1. **At \( x = \frac{\pi}{4} \)**:
\[ f\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{1 + \tan\left(\frac{\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{1 + 1} = \frac{\frac{\sqrt{2}}{2}}{2} = \frac{\sqrt{2}}{4} \]

2. **At \( x = \frac{5\pi}{4} \)**:
\[ f\left(\frac{5\pi}{4}\right) = \frac{\sin\left(\frac{5\pi}{4}\right)}{1 + \tan\left(\frac{5\pi}{4}\right)} = \frac{-\frac{\sqrt{2}}{2}}{1 - 1} = \text{undefined} \]

### Conclusion
The maximum value of \( f(x) \) occurs at \( x = \frac{\pi}{4} \) with \( f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{4} \), and the minimum value occurs at \( x = \frac{3\pi}{2} \) and \( x = \frac{\pi}{2} \) with \( f\left(\frac{3\pi}{2}\right) = 0 \).