Given the equation:

\[
\sec A + \tan A = 2 + \sqrt{3}
\]

We can use the following trigonometric identities:

\[
\sec A = \frac{1}{\cos A} \quad \text{and} \quad \tan A = \frac{\sin A}{\cos A}
\]

Substituting these into the equation:

\[
\frac{1}{\cos A} + \frac{\sin A}{\cos A} = 2 + \sqrt{3}
\]

Simplify the left-hand side:

\[
\frac{1 + \sin A}{\cos A} = 2 + \sqrt{3}
\]

Now multiply both sides of the equation by \(\cos A\):

\[
1 + \sin A = (2 + \sqrt{3}) \cos A
\]

At this point, the equation becomes quite challenging to solve algebraically without further insight into potential values of \(A\). However, notice that \(\tan A = \sqrt{3}\) holds at a specific value of \(A\), which is often associated with special angles.

Let's try \(A = 60^\circ\) (or \(\frac{\pi}{3}\) radians).

For \(A = 60^\circ\):

\[
\sec 60^\circ = 2 \quad \text{and} \quad \tan 60^\circ = \sqrt{3}
\]

Substituting into the original equation:

\[
\sec 60^\circ + \tan 60^\circ = 2 + \sqrt{3}
\]

This is true, so:

\[
A = 60^\circ \quad \text{or} \quad A = \frac{\pi}{3} \text{ radians}.
\]

Would you like a detailed breakdown or have any further questions?

Here are 5 related questions to explore further:
1. What are the values of \(\sec A\) and \(\tan A\) for other standard angles?
2. Can you solve \(\sec A + \tan A = 3\) using the same method?
3. What are the domain and range of the \(\sec A\) and \(\tan A\) functions?
4. How would this equation change if we added a different constant on the right-hand side?
5. How can you express \(\sec A\) and \(\tan A\) in terms of \(\cos A\) and \(\sin A\)?

**Tip:** Memorizing special angles like \(30^\circ\), \(45^\circ\), and \(60^\circ\) is very useful in solving trigonometric equations quickly.

Solve the trigonometric equation secA + tanA = 2 + √3 using trigonometric identities and special angle theorems. This detailed guide is perfect for students in Grades 9-12 looking to explore concepts like secant, tangent, and special angles such as 60 degrees.

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