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

Given:
\[ \tan\theta + \frac{1}{\tan\theta} = 2 \]

Let \( x = \tan\theta \).
Then the equation becomes:
\[ x + \frac{1}{x} = 2 \]

Step 1: Square both sides
\[ \left(x + \frac{1}{x}\right)^2 = 2^2 \] Expand LHS:
\[ x^2 + \frac{1}{x^2} + 2 = 4 \]
Subtract 2 from both sides:
\[ x^2 + \frac{1}{x^2} = 2 \]

Final Answer:
\[ \boxed{\tan^2\theta + \frac{1}{\tan^2\theta} = 2} \]