In `Delta ABC`, right angled at `B`. If `tan A= (1)/(sqrt3)`, find the value of `sin A. cos C+ cos A sin C`.

Answer

To solve the problem step by step, we will find the value of \( \sin A \cos C + \cos A \sin C \) given that \( \tan A = \frac{1}{\sqrt{3}} \) in triangle \( ABC \) which is right-angled at \( B \).

### Step 1: Understanding the Triangle
We have triangle \( ABC \) with a right angle at \( B \). Since \( \tan A = \frac{1}{\sqrt{3}} \), we can interpret this in terms of the sides of the triangle.

### Step 2: Setting Up the Sides
From the definition of tangent:
\[ \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \]
Let \( AB = K \) (the opposite side) and \( BC = \sqrt{3}K \) (the adjacent side).

### Step 3: Finding the Hypotenuse
Using the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the values:
\[ AC^2 = K^2 + (\sqrt{3}K)^2 = K^2 + 3K^2 = 4K^2 \]
Thus,
\[ AC = 2K \]

### Step 4: Finding \( \sin A \)
Using the definition of sine:
\[ \sin A = \frac{AB}{AC} = \frac{K}{2K} = \frac{1}{2} \]

### Step 5: Finding \( \cos A \)
Using the definition of cosine:
\[ \cos A = \frac{BC}{AC} = \frac{\sqrt{3}K}{2K} = \frac{\sqrt{3}}{2} \]

### Step 6: Finding \( \sin C \)
Since \( C = 90^\circ - A \), we can use the complementary angle identity:
\[ \sin C = \cos A = \frac{\sqrt{3}}{2} \]

### Step 7: Finding \( \cos C \)
Using the complementary angle identity:
\[ \cos C = \sin A = \frac{1}{2} \]

### Step 8: Calculating \( \sin A \cos C + \cos A \sin C \)
Now we can substitute the values we found:
\[ \sin A \cos C + \cos A \sin C = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \]
Calculating each term:
\[ = \frac{1}{4} + \frac{3}{4} = 1 \]

### Final Answer
Thus, the value of \( \sin A \cos C + \cos A \sin C \) is \( 1 \).

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