Find `tan15^(@).tan25^(@).tan45^(@).tan65^(@).tan75^(@)`.

Answer

To find the product \( \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 45^\circ \cdot \tan 65^\circ \cdot \tan 75^\circ \), we can use some trigonometric identities and properties.

### Step-by-Step Solution:

1. **Identify the angles**: We have the angles \( 15^\circ, 25^\circ, 45^\circ, 65^\circ, \) and \( 75^\circ \).

2. **Use the complementary angle identity**:
- We know that \( \tan(90^\circ - \theta) = \cot(\theta) \).
- Therefore, we can express \( \tan 65^\circ \) and \( \tan 75^\circ \) in terms of cotangent:
- \( \tan 65^\circ = \cot 25^\circ \) (since \( 65^\circ = 90^\circ - 25^\circ \))
- \( \tan 75^\circ = \cot 15^\circ \) (since \( 75^\circ = 90^\circ - 15^\circ \))

3. **Substitute these identities into the product**:
\[ \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 45^\circ \cdot \tan 65^\circ \cdot \tan 75^\circ = \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 45^\circ \cdot \cot 25^\circ \cdot \cot 15^\circ \]

4. **Use the identity \( \cot \theta = \frac{1}{\tan \theta} \)**:
- Thus, we can rewrite the product:
\[ = \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 45^\circ \cdot \frac{1}{\tan 25^\circ} \cdot \frac{1}{\tan 15^\circ} \]

5. **Simplify the expression**:
- The \( \tan 25^\circ \) and \( \tan 15^\circ \) cancel out:
\[ = \tan 45^\circ \]

6. **Evaluate \( \tan 45^\circ \)**:
- We know that \( \tan 45^\circ = 1 \).

7. **Final answer**:
\[ \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 45^\circ \cdot \tan 65^\circ \cdot \tan 75^\circ = 1 \]