Find the value of
`tan^(n)1^(@)tan^(n)2^(@)tan^(n)3^(@)......tan^(n)88^(@)tan^(n)89^(@)`.

Options

A. a) 0
B. b) 3
C. c) 1
D. d) 4

Answer

To find the value of \( \tan^n(1^\circ) \tan^n(2^\circ) \tan^n(3^\circ) \ldots \tan^n(88^\circ) \tan^n(89^\circ) \), we can use the property of tangent and cotangent.

### Step-by-Step Solution:

1. **Pairing the Angles**:
Notice that \( \tan(90^\circ - \theta) = \cot(\theta) \). This means that for every angle \( \theta \) from \( 1^\circ \) to \( 89^\circ \), we can pair it with \( 90^\circ - \theta \).
- For example:
- \( \tan(1^\circ) \) pairs with \( \tan(89^\circ) \)
- \( \tan(2^\circ) \) pairs with \( \tan(88^\circ) \)
- \( \tan(3^\circ) \) pairs with \( \tan(87^\circ) \)
- And so on, until \( \tan(44^\circ) \) pairs with \( \tan(46^\circ) \), leaving \( \tan(45^\circ) \) unpaired.

2. **Writing the Product**:
We can express the product as:
\[ \tan^n(1^\circ) \tan^n(89^\circ) \cdot \tan^n(2^\circ) \tan^n(88^\circ) \cdots \tan^n(44^\circ) \tan^n(46^\circ) \cdot \tan^n(45^\circ) \]

3. **Using the Pairing Property**:
Each pair \( \tan^n(k^\circ) \tan^n((90-k)^\circ) \) can be rewritten using the cotangent:
\[ \tan^n(k^\circ) \tan^n(90^\circ - k^\circ) = \tan^n(k^\circ) \cot^n(k^\circ) = \tan^n(k^\circ) \cdot \frac{1}{\tan^n(k^\circ)} = 1 \]

4. **Calculating the Total Product**:
Since each of the pairs contributes a product of 1, we can simplify the entire expression:
\[ \tan^n(1^\circ) \tan^n(89^\circ) = 1, \quad \tan^n(2^\circ) \tan^n(88^\circ) = 1, \quad \ldots, \quad \tan^n(44^\circ) \tan^n(46^\circ) = 1 \]
The only angle left is \( \tan^n(45^\circ) \).

5. **Final Value**:
We know that \( \tan(45^\circ) = 1 \). Therefore,
\[ \tan^n(45^\circ) = 1^n = 1 \]

Thus, the final value of the product \( \tan^n(1^\circ) \tan^n(2^\circ) \tan^n(3^\circ) \ldots \tan^n(88^\circ) \tan^n(89^\circ) \) is:

\[ \boxed{1} \]