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The problem you provided asks to find the value of \((\sin A + \cos A) \sec A\) given that \(\tan A = \frac{5}{12}\).

### Solution:

Given: \(\tan A = \frac{5}{12}\), we can find \(\sin A\) and \(\cos A\) by using the Pythagorean identity.

\[
\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}
\]

This means we can form a right triangle with opposite side 5 and adjacent side 12. Using the Pythagorean theorem to find the hypotenuse \(h\):

\[
h = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]

Now, we can determine \(\sin A\), \(\cos A\), and \(\sec A\):

\[
\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}, \quad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}
\]

\[
\sec A = \frac{1}{\cos A} = \frac{13}{12}
\]

### Now, calculate \((\sin A + \cos A) \sec A\):

\[
\sin A + \cos A = \frac{5}{13} + \frac{12}{13} = \frac{17}{13}
\]

Multiply by \(\sec A = \frac{13}{12}\):

\[
(\sin A + \cos A) \sec A = \frac{17}{13} \times \frac{13}{12} = \frac{17}{12}
\]

Thus, the value of \((\sin A + \cos A) \sec A\) is \(\frac{17}{12}\).

---

Would you like more details on any part of this solution? Here are five related questions to help you explore further:

1. How would you apply the same process if \(\tan A = \frac{7}{24}\)?
2. What is the general formula for \(\sec A\) in terms of \(\tan A\)?
3. How does the Pythagorean theorem help in trigonometric identities?
4. Can you derive \(\sec A\) using only the value of \(\tan A\)?
5. How would this problem change if \(\sin A\) and \(\cos A\) were negative?

### Tip: 
Always check if a trigonometric value falls in a specific quadrant, as it can affect the signs of \(\sin A\), \(\cos A\), and \(\sec A\).

If tan A = 5/12, find the value of (sin A + cos A) sec A.

Learn how to find the value of (sin A + cos A) sec A given that tan A = 5/12. This solution explains the use of trigonometric ratios, Pythagorean identities, and key formulas in a step-by-step approach. Ideal for students in Grades 9-10.

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