In `DeltaABC,angleC=90^(@)andtanA=(1)/(sqrt(3))`,find the value of sin A cosB+cosAsinB.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have a right triangle \( \Delta ABC \) with \( \angle C = 90^\circ \) and \( \tan A = \frac{1}{\sqrt{3}} \). We need to find the value of \( \sin A \cos B + \cos A \sin B \).

### Step 2: Draw the triangle
Let's label the triangle:
- \( C \) is the right angle.
- Let \( A \) be one of the angles, and \( B \) be the other angle.

### Step 3: Use the tangent value to find the sides
Given \( \tan A = \frac{1}{\sqrt{3}} \), we can interpret this as:
\[ \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} \]
Let the opposite side (to angle \( A \)) be \( 1 \) and the adjacent side be \( \sqrt{3} \).

### Step 4: Find the hypotenuse
Using the Pythagorean theorem:
\[ \text{hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 5: Find \( \sin A \) and \( \cos A \)
Now we can find \( \sin A \) and \( \cos A \):
\[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2} \]
\[ \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \]

### Step 6: Find \( \sin B \) and \( \cos B \)
Since \( A + B = 90^\circ \), we have:
\[ \sin B = \cos A = \frac{\sqrt{3}}{2} \]
\[ \cos B = \sin A = \frac{1}{2} \]

### Step 7: Substitute into the expression
Now we substitute into the expression \( \sin A \cos B + \cos A \sin B \):
\[ \sin A \cos B + \cos A \sin B = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \]
\[ = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1 \]

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

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