(c) Find the equation of the tangent to the curve \( y=\sec x \) at the point where \( x=\frac{\pi}{4} \)

To find the equation of the tangent to the curve \( y = \sec x \) at the point where \( x = \frac{\pi}{4} \), we need to follow these steps: 1. Find the derivative of the function \( y = \sec x \). 2. Evaluate the derivative at \( x = \frac{\pi}{4} \) to find the slope of the tangent line. 3. Find the y-coordinate of the point on the curve where \( x = \frac{\pi}{4} \). 4. Use the point-slope form of a line to write the equation of the tangent line. Step 1: Find the derivative of \( y = \sec x \). The derivative of \( \sec x \) is \( \sec x \tan x \) because the derivative of \( \sec x \) is \( \sec x \tan x \) according to the derivative rules. Step 2: Evaluate the derivative at \( x = \frac{\pi}{4} \). The derivative at \( x = \frac{\pi}{4} \) is \( \sec \left( \frac{\pi}{4} \right) \tan \left( \frac{\pi}{4} \right) \). Since \( \sec \left( \frac{\pi}{4} \right) = \frac{1}{\cos \left( \frac{\pi}{4} \right)} \) and \( \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), we have: \[ \sec \left( \frac{\pi}{4} \right) = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \] And since \( \tan \left( \frac{\pi}{4} \right) = \frac{\sin \left( \frac{\pi}{4} \right)}{\cos \left( \frac{\pi}{4} \right)} \) and \( \sin \left( \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), we have: \[ \tan \left( \frac{\pi}{4} \right) = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1 \] So the slope of the tangent line is: \[ \sqrt{2} \cdot 1 = \sqrt{2} \] Step 3: Find the y-coordinate of the point on the curve where \( x = \frac{\pi}{4} \). To find the y-coordinate, we substitute \( x = \frac{\pi}{4} \) into the original function: \[ y = \sec \left( \frac{\pi}{4} \right) = \sqrt{2} \] So the point is \( \left( \frac{\pi}{4}, \sqrt{2} \right) \). Step 4: Use the point-slope form of a line to write the equation of the tangent line. The point-slope form of a line is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. We have \( m = \sqrt{2} \) and \( (x_1, y_1) = \left( \frac{\pi}{4}, \sqrt{2} \right) \). Plugging these values into the point-slope form, we get: \[ y - \sqrt{2} = \sqrt{2} \left( x - \frac{\pi}{4} \right) \] This is the equation of the tangent to the curve \( y = \sec x \) at the point where \( x = \frac{\pi}{4} \).