Find the equations of the tangent and normal lines to the following curves :
`y=(1+sinx)/(cosx)" at "x=(pi)/(4).`

Answer

To find the equations of the tangent and normal lines to the curve \( y = \frac{1 + \sin x}{\cos x} \) at \( x = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Find the point on the curve at \( x = \frac{\pi}{4} \)

First, we need to calculate the value of \( y \) at \( x = \frac{\pi}{4} \):

\[ y = \frac{1 + \sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} \]

Using the known values \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \) and \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):

\[ y = \frac{1 + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{2} + 1}{\frac{1}{\sqrt{2}}} = \sqrt{2}(\sqrt{2} + 1) = 1 + \sqrt{2} \]

Thus, the point on the curve is \( \left(\frac{\pi}{4}, 1 + \sqrt{2}\right) \).

### Step 2: Find the derivative \( \frac{dy}{dx} \)

Next, we need to find the derivative of \( y \) to determine the slope of the tangent line:

\[ y = \frac{1 + \sin x}{\cos x} = \sec x + \tan x \]

Differentiating \( y \):

\[ \frac{dy}{dx} = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) = \sec x \tan x + \sec^2 x \]

### Step 3: Evaluate the derivative at \( x = \frac{\pi}{4} \)

Now we substitute \( x = \frac{\pi}{4} \) into the derivative:

\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{4}} = \sec\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) + \sec^2\left(\frac{\pi}{4}\right) \]

Using the known values \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \) and \( \tan\left(\frac{\pi}{4}\right) = 1 \):

\[ = \sqrt{2} \cdot 1 + (\sqrt{2})^2 = \sqrt{2} + 2 = 2 + \sqrt{2} \]

Thus, the slope of the tangent line at \( x = \frac{\pi}{4} \) is \( 2 + \sqrt{2} \).

### Step 4: Write the equation of the tangent line

Using the point-slope form of the equation of a line:

\[ y - y_1 = m(x - x_1) \]

Where \( (x_1, y_1) = \left(\frac{\pi}{4}, 1 + \sqrt{2}\right) \) and \( m = 2 + \sqrt{2} \):

\[ y - (1 + \sqrt{2}) = (2 + \sqrt{2})\left(x - \frac{\pi}{4}\right) \]

This is the equation of the tangent line.

### Step 5: Write the equation of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line:

\[ m_{\text{normal}} = -\frac{1}{2 + \sqrt{2}} \]

Using the point-slope form again:

\[ y - (1 + \sqrt{2}) = -\frac{1}{2 + \sqrt{2}}\left(x - \frac{\pi}{4}\right) \]

This is the equation of the normal line.

### Final Equations

1. **Equation of the Tangent Line:**
\[ y - (1 + \sqrt{2}) = (2 + \sqrt{2})\left(x - \frac{\pi}{4}\right) \]

2. **Equation of the Normal Line:**
\[ y - (1 + \sqrt{2}) = -\frac{1}{2 + \sqrt{2}}\left(x - \frac{\pi}{4}\right) \]