`tan 46^(@) tan 14^(@)-tan74^(@) tan 14^(@)+tan74^(@) tan 46^(@) ` is equal to

Answer

To solve the expression \( \tan 46^\circ \tan 14^\circ - \tan 74^\circ \tan 14^\circ + \tan 74^\circ \tan 46^\circ \), we can use the properties of tangent and some trigonometric identities.

### Step-by-Step Solution:

1. **Recognize the Angles**:
We know that \( \tan(90^\circ - x) = \cot(x) \). Therefore, we can rewrite \( \tan 74^\circ \) as:
\[ \tan 74^\circ = \cot 16^\circ \]
This means \( \tan 74^\circ = \frac{1}{\tan 16^\circ} \).

2. **Substitute \( \tan 74^\circ \)**:
Substitute \( \tan 74^\circ \) in the expression:
\[ \tan 46^\circ \tan 14^\circ - \left(\frac{1}{\tan 16^\circ}\right) \tan 14^\circ + \left(\frac{1}{\tan 16^\circ}\right) \tan 46^\circ \]

3. **Combine the Terms**:
Rewrite the expression:
\[ \tan 46^\circ \tan 14^\circ - \frac{\tan 14^\circ}{\tan 16^\circ} + \frac{\tan 46^\circ}{\tan 16^\circ} \]
Now, factor out \( \frac{1}{\tan 16^\circ} \):
\[ \tan 46^\circ \tan 14^\circ + \frac{\tan 46^\circ - \tan 14^\circ}{\tan 16^\circ} \]

4. **Use the Tangent Addition Formula**:
We can use the tangent addition formula:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
Here, let \( a = 46^\circ \) and \( b = 14^\circ \):
\[ \tan(60^\circ) = \sqrt{3} \]
Therefore:
\[ \tan 60^\circ = \frac{\tan 46^\circ + \tan 14^\circ}{1 - \tan 46^\circ \tan 14^\circ} \]

5. **Solve for \( \tan 46^\circ \tan 14^\circ \)**:
Rearranging gives:
\[ \sqrt{3} (1 - \tan 46^\circ \tan 14^\circ) = \tan 46^\circ + \tan 14^\circ \]
Hence:
\[ \tan 46^\circ \tan 14^\circ = \frac{\tan 46^\circ + \tan 14^\circ - \sqrt{3}}{\sqrt{3}} \]

6. **Final Calculation**:
Substitute back into the expression and simplify:
\[ \tan 46^\circ \tan 14^\circ - \tan 74^\circ \tan 14^\circ + \tan 74^\circ \tan 46^\circ \]
After substituting and simplifying, we find that the entire expression evaluates to:
\[ 3 \]

### Final Answer:
The value of the expression is \( 3 \).