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We are given that \( \cot A = 7 \), and we need to find the value of the expression:

\[
\frac{5 \cos A + 4 \sin A}{\cos^3 A + 7 \sin^3 A + 6 \sin A}
\]

### Step 1: Express \(\sin A\) and \(\cos A\) in terms of \(\cot A\)
We know:
\[
\cot A = \frac{\cos A}{\sin A}
\]
Given that \( \cot A = 7 \), we have:
\[
\frac{\cos A}{\sin A} = 7 \implies \cos A = 7 \sin A
\]

Now, using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \), substitute \( \cos A = 7 \sin A \) into this identity:
\[
\sin^2 A + (7 \sin A)^2 = 1
\]
\[
\sin^2 A + 49 \sin^2 A = 1
\]
\[
50 \sin^2 A = 1 \implies \sin^2 A = \frac{1}{50}
\]
\[
\sin A = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}}
\]

Now, use \( \cos A = 7 \sin A \) to find \( \cos A \):
\[
\cos A = 7 \times \frac{1}{5\sqrt{2}} = \frac{7}{5\sqrt{2}}
\]

### Step 2: Substitute values into the expression
Now that we have \( \sin A = \frac{1}{5\sqrt{2}} \) and \( \cos A = \frac{7}{5\sqrt{2}} \), we can substitute these into the expression:

\[
\frac{5 \cos A + 4 \sin A}{\cos^3 A + 7 \sin^3 A + 6 \sin A}
\]

#### Numerator:
\[
5 \cos A + 4 \sin A = 5 \times \frac{7}{5\sqrt{2}} + 4 \times \frac{1}{5\sqrt{2}} = \frac{35}{5\sqrt{2}} + \frac{4}{5\sqrt{2}} = \frac{39}{5\sqrt{2}}
\]

#### Denominator:
\[
\cos^3 A = \left( \frac{7}{5\sqrt{2}} \right)^3 = \frac{343}{125 \cdot \sqrt{8}} = \frac{343}{500\sqrt{2}}
\]
\[
\sin^3 A = \left( \frac{1}{5\sqrt{2}} \right)^3 = \frac{1}{125 \cdot \sqrt{8}} = \frac{1}{500\sqrt{2}}
\]
Now, calculate the entire denominator:
\[
\cos^3 A + 7 \sin^3 A + 6 \sin A = \frac{343}{500\sqrt{2}} + 7 \times \frac{1}{500\sqrt{2}} + 6 \times \frac{1}{5\sqrt{2}}
\]
\[
= \frac{343}{500\sqrt{2}} + \frac{7}{500\sqrt{2}} + \frac{6}{5\sqrt{2}} = \frac{350}{500\sqrt{2}} + \frac{6}{5\sqrt{2}} = \frac{350}{500\sqrt{2}} + \frac{600}{500\sqrt{2}}
\]
\[
= \frac{950}{500\sqrt{2}} = \frac{19}{10\sqrt{2}}
\]

### Step 3: Final Simplification
Now, we can substitute these into the main expression:
\[
\frac{\frac{39}{5\sqrt{2}}}{\frac{19}{10\sqrt{2}}} = \frac{39}{5\sqrt{2}} \times \frac{10\sqrt{2}}{19} = \frac{390}{95} = \frac{78}{19}
\]

Thus, the value of the expression is:
\[
\boxed{\frac{78}{19}}
\]

Would you like further details or explanations on any part of the solution?

---

### 5 Related Questions:
1. How do we use the Pythagorean identity to express \( \sin A \) and \( \cos A \) in terms of \( \cot A \)?
2. What are the steps to simplify a trigonometric fraction when given \( \cot A \)?
3. How do you calculate cube powers of trigonometric functions like \( \cos^3 A \) and \( \sin^3 A \)?
4. How can you use identities like \( \cos^2 A + \sin^2 A = 1 \) to solve for unknowns?
5. What is the importance of rationalizing trigonometric expressions involving square roots?

### Tip:
Whenever working with trigonometric identities, always simplify using basic identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) before substituting into more complex expressions!

If cot A = 7, then the value of (5 cos A + 4 sin A) / (cos^3 A + 7 sin^3 A + 6 sin A) is:

This problem explores the calculation of the trigonometric expression (5 cos A + 4 sin A) / (cos^3 A + 7 sin^3 A + 6 sin A) given that cot A = 7. It involves the use of trigonometric identities and algebraic simplification. Suitable for students in Grades 10-12.

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