Question

If \( \tan \theta + \frac{1}{\tan \theta} = 2 \), find the value of \( \tan^2 \theta + \frac{1}{\tan^2 \theta} \).

Hint:

If \( x + \frac{1}{x} = 2 \), then \( x^n + \frac{1}{x^n} \) is always 2 for any natural number \( n \).

Answer 1

Step 1: Understanding the Concept:
We use the algebraic identity \( (a + b)^2 = a^2 + b^2 + 2ab \).
Step 2: Detailed Explanation:
Given: \( \tan \theta + \frac{1}{\tan \theta} = 2 \).
Squaring both sides:
\[ \left( \tan \theta + \frac{1}{\tan \theta} \right)^2 = (2)^2 \]
\[ \tan^2 \theta + \left( \frac{1}{\tan \theta} \right)^2 + 2(\tan \theta) \left( \frac{1}{\tan \theta} \right) = 4 \]
Since \( \tan \theta \times \frac{1}{\tan \theta} = 1 \):
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} + 2 = 4 \]
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} = 4 - 2 = 2 \]
Step 3: Final Answer:
The value is 2.