If `tan theta+cot theta=2` then value of `tan^(n) theta+ cot^(n) theta(0^(@) lt theta lt 90^(@)` n is an integar) is
यदि `tan theta+cot theta=2`, तो `tan^(n) theta+cot^(n) (0^(@) lt theta lt 90^(@)` एक पूर्णाक है) का मान है

Options

A. 2
B. `2^(n)`
C. 2n
D. `2^(n+1)`

Explanation

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

1. **Given Equation**:
\[ \tan \theta + \cot \theta = 2 \]

2. **Using the Identity**:
We know that \(\tan \theta\) can be expressed as \(\frac{1}{\cot \theta}\). Therefore, we can rewrite the equation as:
\[ \tan \theta + \frac{1}{\tan \theta} = 2 \]

3. **Letting \(x = \tan \theta\)**:
We can substitute \(x\) for \(\tan \theta\):
\[ x + \frac{1}{x} = 2 \]

4. **Multiplying through by \(x\)**:
To eliminate the fraction, multiply the entire equation by \(x\):
\[ x^2 + 1 = 2x \]

5. **Rearranging the Equation**:
Rearranging gives us a standard quadratic equation:
\[ x^2 - 2x + 1 = 0 \]

6. **Factoring the Quadratic**:
This can be factored as:
\[ (x - 1)^2 = 0 \]

7. **Finding the Value of \(x\)**:
Solving for \(x\) gives:
\[ x - 1 = 0 \implies x = 1 \]
Thus, \(\tan \theta = 1\) and consequently \(\cot \theta = 1\).

8. **Calculating \(\tan^n \theta + \cot^n \theta\)**:
Now we need to find:
\[ \tan^n \theta + \cot^n \theta = 1^n + 1^n \]

9. **Simplifying the Expression**:
Since \(1^n = 1\) for any integer \(n\):
\[ \tan^n \theta + \cot^n \theta = 1 + 1 = 2 \]

10. **Final Answer**:
Therefore, the value of \(\tan^n \theta + \cot^n \theta\) is:
\[ \boxed{2} \]