If `tan alpha+cotalpha=2`, then `tan^(20)alpha+cot^(20)alpha` =

Options

A. `0`
B. `2`
C. `20`
D. `2^(20)`

Answer

To solve the problem, we need to find the value of \( \tan^{20} \alpha + \cot^{20} \alpha \) given that \( \tan \alpha + \cot \alpha = 2 \).

### Step 1: Understand the given equation
We start with the equation:
\[ \tan \alpha + \cot \alpha = 2 \]

### Step 2: Substitute values for \( \tan \alpha \) and \( \cot \alpha \)
Recall that \( \cot \alpha = \frac{1}{\tan \alpha} \). Let \( x = \tan \alpha \). Then we can rewrite the equation as:
\[ x + \frac{1}{x} = 2 \]

### Step 3: Multiply through by \( x \)
To eliminate the fraction, multiply both sides by \( x \):
\[ x^2 + 1 = 2x \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ x^2 - 2x + 1 = 0 \]

### Step 5: Factor the quadratic equation
This can be factored as:
\[ (x - 1)^2 = 0 \]

### Step 6: Solve for \( x \)
Setting the factor equal to zero gives:
\[ x - 1 = 0 \implies x = 1 \]
Thus, \( \tan \alpha = 1 \) and consequently \( \cot \alpha = 1 \).

### Step 7: Calculate \( \tan^{20} \alpha + \cot^{20} \alpha \)
Now we can find:
\[ \tan^{20} \alpha + \cot^{20} \alpha = 1^{20} + 1^{20} = 1 + 1 = 2 \]

### Final Answer
Thus, we conclude that:
\[ \tan^{20} \alpha + \cot^{20} \alpha = 2 \]

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