Find `tan1^(@).tan2^(@)……………….tan89^(@)`

Answer

To find the product \( \tan(1^\circ) \tan(2^\circ) \tan(3^\circ) \ldots \tan(89^\circ) \), we can utilize the properties of the tangent function and its relationship with the cotangent function.

### Step-by-Step Solution:

1. **Write the Product**:
\[ P = \tan(1^\circ) \tan(2^\circ) \tan(3^\circ) \ldots \tan(89^\circ) \]

2. **Pairing the Terms**:
Notice that \( \tan(90^\circ - \theta) = \cot(\theta) \). Therefore, we can pair the terms:
\[ \tan(1^\circ) \tan(89^\circ), \tan(2^\circ) \tan(88^\circ), \ldots, \tan(44^\circ) \tan(46^\circ) \]
The middle term is \( \tan(45^\circ) \).

3. **Using the Identity**:
Each pair can be expressed as:
\[ \tan(k^\circ) \tan(90^\circ - k^\circ) = \tan(k^\circ) \cot(k^\circ) = 1 \]
Thus, we have:
\[ P = (\tan(1^\circ) \tan(89^\circ)) \cdot (\tan(2^\circ) \tan(88^\circ)) \cdots (\tan(44^\circ) \tan(46^\circ)) \cdot \tan(45^\circ) \]

4. **Calculating the Product**:
Since each of the pairs equals 1, we have:
\[ P = 1 \cdot 1 \cdot 1 \cdots \cdot \tan(45^\circ) \]
And we know that:
\[ \tan(45^\circ) = 1 \]

5. **Final Result**:
Therefore, the entire product simplifies to:
\[ P = 1 \]

### Conclusion:
The value of \( \tan(1^\circ) \tan(2^\circ) \ldots \tan(89^\circ) \) is \( 1 \).