If `tantheta+cottheta=5` , then `tan^(2)theta+cot^(2)theta` is

Options

A. 23
B. 25
C. 26
D. 24

Answer

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

1. **Given:**
\[ \tan \theta + \cot \theta = 5 \]

2. **Using the identity:**
We know that:
\[ \tan \theta \cdot \cot \theta = 1 \]
Let \( x = \tan \theta \). Then, we can express \( \cot \theta \) as:
\[ \cot \theta = \frac{1}{x} \]
Substituting this into the original equation gives:
\[ x + \frac{1}{x} = 5 \]

3. **Multiplying through by \( x \):**
To eliminate the fraction, multiply both sides by \( x \):
\[ x^2 + 1 = 5x \]

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

5. **Using the quadratic formula:**
We can solve for \( x \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -5 \), and \( c = 1 \):
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{5 \pm \sqrt{25 - 4}}{2} \]
\[ x = \frac{5 \pm \sqrt{21}}{2} \]

6. **Calculating \( \tan^2 \theta + \cot^2 \theta \):**
We need to find \( \tan^2 \theta + \cot^2 \theta \). We can use the identity:
\[ \tan^2 \theta + \cot^2 \theta = (\tan \theta + \cot \theta)^2 - 2 \]
Substituting the known value:
\[ \tan^2 \theta + \cot^2 \theta = 5^2 - 2 \]
\[ \tan^2 \theta + \cot^2 \theta = 25 - 2 = 23 \]

7. **Final answer:**
Therefore, the value of \( \tan^2 \theta + \cot^2 \theta \) is:
\[ \boxed{23} \]