https://file.aiquickdraw.com/mathbot/file/1726410348390gdh7kb9k.jpg

The equation given in the image is:

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

We can simplify the left-hand side:

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

Thus, the equation becomes:

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

Now, we need to solve for \( A \). One approach is to recall standard values of \( \sec A + \tan A \) for common angles. We can try to match the right-hand side with known trigonometric values.

For \( A = 30^\circ \) or \( A = \frac{\pi}{6} \):

\[
\sec 30^\circ = \frac{2}{\sqrt{3}}, \quad \tan 30^\circ = \frac{1}{\sqrt{3}}
\]
\[
\sec 30^\circ + \tan 30^\circ = \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}
\]

For \( A = 60^\circ \) or \( A = \frac{\pi}{3} \):

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

This matches the right-hand side of the equation exactly. Therefore, the value of \( A \) is:

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

Would you like more details on any specific step or further explanation?

Here are 5 related questions to explore further:
1. What are the definitions of secant and tangent in terms of sine and cosine?
2. How can you derive exact trigonometric values for standard angles like \( 30^\circ \) and \( 60^\circ \)?
3. What is the geometric meaning of secant and tangent?
4. Can this equation be solved graphically to verify the result?
5. How do secant and tangent behave as \( A \) approaches 90 degrees?

**Tip**: Knowing standard trigonometric values for common angles like 30°, 45°, and 60° is very useful for solving equations quickly!

If 1/cosA + sinA/cosA = 2 + √3, what is the value of A?

Learn how to solve the trigonometric equation 1/cosA + sinA/cosA = 2 + √3 and find the value of A using secant and tangent identities. This problem covers key trigonometric concepts and simplifications, suitable for Grades 10-12.

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