If `tan x + cot x =2` , then `sin^(2n)x + cos^(2n)`x is equal to

Options

A. `2^n`
B. `- 1/2`
C. `1/2`
D. None of these

Answer

To solve the problem where \( \tan x + \cot x = 2 \), we need to find the value of \( \sin^{2n} x + \cos^{2n} x \).

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We start with the equation:
\[ \tan x + \cot x = 2 \]
We know that \( \tan x = \frac{\sin x}{\cos x} \) and \( \cot x = \frac{\cos x}{\sin x} \).

2. **Using AM-GM Inequality**:
Since both \( \tan x \) and \( \cot x \) are positive (as \( \tan x + \cot x = 2 \) implies both are greater than 0), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{\tan x + \cot x}{2} \geq \sqrt{\tan x \cdot \cot x} \]
This simplifies to:
\[ \frac{\tan x + \cot x}{2} \geq 1 \]
Therefore, we have:
\[ \tan x + \cot x \geq 2 \]

3. **Equality Condition**:
The equality in AM-GM holds when:
\[ \tan x = \cot x \]
This implies:
\[ \tan^2 x = 1 \quad \Rightarrow \quad \tan x = 1 \]
Thus, we find:
\[ x = \frac{\pi}{4} + k\pi \quad (k \in \mathbb{Z}) \]

4. **Finding \( \sin x \) and \( \cos x \)**:
For \( x = \frac{\pi}{4} \):
\[ \sin x = \cos x = \frac{1}{\sqrt{2}} \]

5. **Calculating \( \sin^{2n} x + \cos^{2n} x \)**:
Now we substitute \( x = \frac{\pi}{4} \) into the expression:
\[ \sin^{2n} x + \cos^{2n} x = \left(\frac{1}{\sqrt{2}}\right)^{2n} + \left(\frac{1}{\sqrt{2}}\right)^{2n} \]
This simplifies to:
\[ 2 \left(\frac{1}{\sqrt{2}}\right)^{2n} = 2 \left(\frac{1}{2}\right)^{n} = 2^{1-n} \]

6. **Final Result**:
Therefore, we conclude that:
\[ \sin^{2n} x + \cos^{2n} x = 2^{1-n} \]

### Conclusion:
The value of \( \sin^{2n} x + \cos^{2n} x \) is \( 2^{1-n} \).