If `tan alpha+cot alpha=sqrt3`, then `tan^(3)alpha+cot^(3)alpha` is equal to

Options

A. 0
B. `sqrt3`
C. `3sqrt3`
D. 1

Answer

To solve the problem, we need to find the value of \( \tan^3 \alpha + \cot^3 \alpha \) given that \( \tan \alpha + \cot \alpha = \sqrt{3} \).

### Step-by-Step Solution:

1. **Let \( x = \tan \alpha \)**:
- Then, \( \cot \alpha = \frac{1}{x} \).
- Therefore, we can rewrite the equation as:
\[ x + \frac{1}{x} = \sqrt{3} \]

2. **Multiply both sides by \( x \)**:
- This gives us:
\[ x^2 + 1 = \sqrt{3} x \]

3. **Rearranging the equation**:
- We can rearrange it to form a quadratic equation:
\[ x^2 - \sqrt{3} x + 1 = 0 \]

4. **Using the identity for cubes**:
- We know that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
- Here, \( a = \tan \alpha \) and \( b = \cot \alpha \).
- Therefore, we can express \( \tan^3 \alpha + \cot^3 \alpha \) as:
\[ \tan^3 \alpha + \cot^3 \alpha = (\tan \alpha + \cot \alpha)((\tan \alpha)^2 - \tan \alpha \cot \alpha + (\cot \alpha)^2) \]

5. **Calculating \( \tan^2 \alpha + \cot^2 \alpha \)**:
- We can find \( \tan^2 \alpha + \cot^2 \alpha \) using the identity:
\[ \tan^2 \alpha + \cot^2 \alpha = (x + \frac{1}{x})^2 - 2 = (\sqrt{3})^2 - 2 = 3 - 2 = 1 \]

6. **Calculating \( \tan \alpha \cot \alpha \)**:
- We know that:
\[ \tan \alpha \cot \alpha = 1 \]

7. **Putting it all together**:
- Now substituting back into the identity:
\[ \tan^3 \alpha + \cot^3 \alpha = (\tan \alpha + \cot \alpha)((\tan^2 \alpha + \cot^2 \alpha) - \tan \alpha \cot \alpha) \]
- This simplifies to:
\[ \tan^3 \alpha + \cot^3 \alpha = \sqrt{3}(1 - 1) = \sqrt{3} \cdot 0 = 0 \]

### Final Answer:
\[ \tan^3 \alpha + \cot^3 \alpha = 0 \]