If `tanalpha+cotalpha=2`, then

Options

A. (a)`tan^(n)alpha + cot^(n)alpha=2`
B. (b)`sqrt(tanalpha)+sqrt(cotalpha)=2`
C. (c)`sin^(2n)alpha+cos^(2n)alpha=1/(2^(n-1))`
D. (d)`tan^(2)alpha+sec^(2)alpha=3`

Answer

To solve the problem, we start with the given equation:

**Given:**
\[ \tan \alpha + \cot \alpha = 2 \]

**Step 1: Square both sides of the equation.**
\[ (\tan \alpha + \cot \alpha)^2 = 2^2 \]
This expands to:
\[ \tan^2 \alpha + 2 \tan \alpha \cot \alpha + \cot^2 \alpha = 4 \]

**Step 2: Simplify the equation.**
Since \(\tan \alpha \cot \alpha = 1\), we can substitute this into the equation:
\[ \tan^2 \alpha + 2(1) + \cot^2 \alpha = 4 \]
This simplifies to:
\[ \tan^2 \alpha + \cot^2 \alpha + 2 = 4 \]

**Step 3: Rearrange the equation.**
Subtract 2 from both sides:
\[ \tan^2 \alpha + \cot^2 \alpha = 4 - 2 \]
Thus, we have:
\[ \tan^2 \alpha + \cot^2 \alpha = 2 \]

**Step 4: Generalize for \(n\).**
Now, we can analyze the expression \( \tan^n \alpha + \cot^n \alpha \).

For \(n = 1\):
\[ \tan \alpha + \cot \alpha = 2 \]

For \(n = 2\):
\[ \tan^2 \alpha + \cot^2 \alpha = 2 \]

Thus, we can conclude that:
\[ \tan^n \alpha + \cot^n \alpha = 2 \text{ for } n = 1 \text{ and } n = 2 \]

**Final Conclusion:**
The solution to the problem is that \( \tan^n \alpha + \cot^n \alpha = 2 \) for \( n = 1 \) and \( n = 2 \).

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