given the function y=2secx find the equation of the tangent at x = pi/6

To find the equation of the tangent to the function \( y = 2\sec(x) \) at \( x = \frac{\pi}{6} \), follow these steps:

### 1. Differentiate the function:
The equation of the tangent line at a given point is found using the derivative of the function, which gives the slope of the tangent line.

We have \( y = 2\sec(x) \).

The derivative of \( \sec(x) \) with respect to \( x \) is:

\[
\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)
\]

So, differentiating \( y = 2\sec(x) \) with respect to \( x \):

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

This gives the slope of the tangent line at any point \( x \).

### 2. Find the slope at \( x = \frac{\pi}{6} \):
Now, substitute \( x = \frac{\pi}{6} \) into the derivative to find the slope of the tangent line.

At \( x = \frac{\pi}{6} \):
\[
\sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}
\]
\[
\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}
\]

Thus, the slope at \( x = \frac{\pi}{6} \) is:

\[
\frac{dy}{dx}\bigg|_{x = \frac{\pi}{6}} = 2 \cdot \frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = \frac{4}{3}
\]

### 3. Find the point of tangency:
At \( x = \frac{\pi}{6} \), substitute into the original function to find the corresponding \( y \)-value.

\[
y = 2\sec\left(\frac{\pi}{6}\right) = 2 \cdot \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}
\]

So, the point of tangency is \( \left(\frac{\pi}{6}, \frac{4\sqrt{3}}{3}\right) \).

### 4. Write the equation of the tangent line:
The equation of a line is given by:

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

Where \( m \) is the slope, and \( (x_1, y_1) \) is the point of tangency.

Substituting the values \( m = \frac{4}{3} \), \( x_1 = \frac{\pi}{6} \), and \( y_1 = \frac{4\sqrt{3}}{3} \):

\[
y - \frac{4\sqrt{3}}{3} = \frac{4}{3}\left(x - \frac{\pi}{6}\right)
\]

This is the equation of the tangent line.

### Final Answer:
The equation of the tangent line to \( y = 2\sec(x) \) at \( x = \frac{\pi}{6} \) is:

\[
y - \frac{4\sqrt{3}}{3} = \frac{4}{3}\left(x - \frac{\pi}{6}\right)
\]

Let me know if you would like further details or have any questions.

Here are some related questions to deepen your understanding:

1. How do you find the equation of the tangent line for a general function at a specific point?
2. What is the difference between secant and tangent functions in terms of calculus and geometry?
3. Can you find the normal line at the same point \( x = \frac{\pi}{6} \)?
4. What is the significance of the derivative in determining the slope of a tangent?
5. How would the equation change if the function were \( y = 3\sec(x) \)?

**Tip**: Always check whether the function has vertical asymptotes near the point of tangency to ensure the tangent exists.

Given the function y=2secx, find the equation of the tangent at x = pi/6.

Find the equation of the tangent line to y = 2sec(x) at x = pi/6 using differentiation. This detailed solution includes trigonometric identities and calculus principles, perfect for students in Grades 11-12 studying trigonometry and calculus.

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