In a triangle ABC,k right angled at B, if `tan A = (1)/(sqrt3).` Find the value of cos A.cosC-sinA.sinC.`

Answer

To solve the problem step by step, we will follow the information given in the question and the video transcript.

### Step 1: Understand the triangle
We are given a triangle ABC, which is right-angled at B. This means that angle B = 90°.

### Step 2: Use the tangent value
We are given that \( \tan A = \frac{1}{\sqrt{3}} \). We know that:
\[ \tan A = \frac{\text{opposite}}{\text{adjacent}} \]
From trigonometric ratios, we can find the angle A. The value of \( \tan 30° = \frac{1}{\sqrt{3}} \). Therefore, we conclude:
\[ A = 30° \]

### Step 3: Find angle C
Since the sum of angles in a triangle is 180°, we can find angle C:
\[ C = 90° - A = 90° - 30° = 60° \]

### Step 4: Find the trigonometric values
Now, we need to find \( \cos A, \cos C, \sin A, \) and \( \sin C \):
- \( \cos A = \cos 30° = \frac{\sqrt{3}}{2} \)
- \( \cos C = \cos 60° = \frac{1}{2} \)
- \( \sin A = \sin 30° = \frac{1}{2} \)
- \( \sin C = \sin 60° = \frac{\sqrt{3}}{2} \)

### Step 5: Substitute into the expression
We need to find the value of \( \cos A \cos C - \sin A \sin C \):
\[ \cos A \cos C - \sin A \sin C = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right) - \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \]

### Step 6: Simplify the expression
Calculating the above expression:
\[ = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 \]

### Final Answer
Thus, the value of \( \cos A \cos C - \sin A \sin C \) is:
\[ \boxed{0} \]

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