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We are given a triangle \( ABC \), right-angled at \( B \), with the information that:

\[
\text{tan} A = \frac{1}{\sqrt{3}}
\]

We are asked to find the value of the following two expressions:

1. \( \sin A \cos C + \cos A \sin C \)
2. \( \cos A \cos C - \sin A \sin C \)

### Step 1: Analyzing the information

Since \( \text{tan} A = \frac{1}{\sqrt{3}} \), we can determine the trigonometric ratios for angle \( A \). Recall that:

\[
\text{tan} A = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}}
\]

This implies that in a right triangle, the opposite side to angle \( A \) is 1, and the adjacent side to \( A \) is \( \sqrt{3} \). Using the Pythagorean theorem, the hypotenuse \( AB \) can be calculated as:

\[
AB = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
\]

Thus, the sides of the triangle are:

- Opposite to \( A \): 1
- Adjacent to \( A \): \( \sqrt{3} \)
- Hypotenuse: 2

### Step 2: Calculating \( \sin A \) and \( \cos A \)

From the triangle:

\[
\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2}
\]
\[
\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}
\]

### Step 3: Relationship between angles \( A \) and \( C \)

Since the triangle is right-angled at \( B \), we know that \( A + C = 90^\circ \). Therefore:

\[
C = 90^\circ - A
\]

This gives us the following relationships for the trigonometric functions of \( C \):

\[
\sin C = \cos A \quad \text{and} \quad \cos C = \sin A
\]

### Step 4: Solving the given expressions

#### Expression 1: \( \sin A \cos C + \cos A \sin C \)

Substituting the values:

\[
\sin A \cos C + \cos A \sin C = \sin A \cdot \sin A + \cos A \cdot \cos A
\]
\[
= \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right) + \left( \frac{\sqrt{3}}{2} \right) \cdot \left( \frac{\sqrt{3}}{2} \right)
\]
\[
= \frac{1}{4} + \frac{3}{4} = 1
\]

Thus, the value of the first expression is:

\[
1
\]

#### Expression 2: \( \cos A \cos C - \sin A \sin C \)

Substituting the values:

\[
\cos A \cos C - \sin A \sin C = \cos A \cdot \sin A - \sin A \cdot \sin A
\]
\[
= \left( \frac{\sqrt{3}}{2} \right) \cdot \left( \frac{1}{2} \right) - \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right)
\]
\[
= \frac{\sqrt{3}}{4} - \frac{1}{4} = \frac{\sqrt{3} - 1}{4}
\]

Thus, the value of the second expression is:

\[
\frac{\sqrt{3} - 1}{4}
\]

### Final Answer:
1. \( \sin A \cos C + \cos A \sin C = 1 \)
2. \( \cos A \cos C - \sin A \sin C = \frac{\sqrt{3} - 1}{4} \)

Let me know if you'd like any further details!

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#### Relative Questions:
1. How can you determine trigonometric ratios using a right triangle?
2. What is the relationship between angles in a right-angled triangle?
3. How do you use the Pythagorean theorem to find the hypotenuse?
4. What is the significance of \( \sin(A + C) = 1 \) in this context?
5. How does the identity for \( \cos(A - C) \) work for complementary angles?

#### Tip:
Remember, in right-angled triangles, complementary angles (like \( A \) and \( C \)) have co-function relationships like \( \sin A = \cos C \) and \( \cos A = \sin C \).

In triangle ABC, right-angled at B, if tan A = 1/sqrt(3), find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C − sin A sin C.

This solution explains how to find the value of sin(A + C) and cos(A − C) in a right-angled triangle where tan A = 1/sqrt(3). Step-by-step process involves using trigonometric identities and the Pythagorean theorem. Suitable for Grades 9-11.

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