If `tantheta+cottheta=2` then find the value of `tan^(2)theta+cot^(2)theta`.

Options

A. 2
B. 1
C. `sqrt(2)`
D. 0

Answer

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

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

2. **Rewrite cotangent**:
We can express cotangent in terms of tangent:
\[ \cot \theta = \frac{1}{\tan \theta} \]
So, substituting this into the equation gives:
\[ \tan \theta + \frac{1}{\tan \theta} = 2 \]

3. **Let \( x = \tan \theta \)**:
This simplifies our equation to:
\[ x + \frac{1}{x} = 2 \]

4. **Multiply through by \( x \)** (to eliminate the fraction):
\[ x^2 + 1 = 2x \]

5. **Rearranging the equation**:
\[ x^2 - 2x + 1 = 0 \]

6. **Factoring the quadratic**:
\[ (x - 1)^2 = 0 \]

7. **Finding \( x \)**:
Thus, we find:
\[ x - 1 = 0 \implies x = 1 \]
Therefore,
\[ \tan \theta = 1 \]

8. **Finding \( \tan^2 \theta + \cot^2 \theta \)**:
Now we need to find \( \tan^2 \theta + \cot^2 \theta \):
\[ \tan^2 \theta = 1^2 = 1 \]
And since \( \cot \theta = \frac{1}{\tan \theta} \):
\[ \cot \theta = 1 \implies \cot^2 \theta = 1^2 = 1 \]

9. **Adding the squares**:
\[ \tan^2 \theta + \cot^2 \theta = 1 + 1 = 2 \]

Thus, the final answer is:
\[ \tan^2 \theta + \cot^2 \theta = 2 \]

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