Question

If \( \alpha = \tan^2 x + \cot^2 x \), where \( x \in \left( 0, \frac{\pi}{2} \right) \), then \( \alpha \) lies in the interval:

• A. \( (-\infty, 1) \)
• B. \( (1, 2) \)
• C. \( (-\infty, 1] \)
• D. \( (-\infty, 2) \)
• E. \( [2, \infty) \)

Hint:

In problems involving trigonometric identities, leveraging relationships between \( \sec x, \csc x, \tan x, \) and \( \cot x \) can simplify complex expressions effectively.

Answer 1

Answer: (E)

We start with the given expression: \[ \alpha = \tan^2 x + \cot^2 x \] This can be transformed using the identity \( \tan x \cdot \cot x = 1 \) to relate \( \tan x \) and \( \cot x \). 
Recognizing that \( \tan^2 x \) and \( \cot^2 x \) are reciprocal squares, we can apply the identity: \[ \tan^2 x + \cot^2 x = \sec^2 x + \csc^2 x - 2 \] Using the pythagorean identities \( \sec^2 x = 1 + \tan^2 x \) and \( \csc^2 x = 1 + \cot^2 x \), we realize: \[ \sec^2 x + \csc^2 x = (1 + \tan^2 x) + (1 + \cot^2 x) = 2 + (\tan^2 x + \cot^2 x) = 2 + \alpha \] Thus, we see: \[ \alpha = \sec^2 x + \csc^2 x - 2 \] The values \( \sec^2 x \) and \( \csc^2 x \) are always greater than or equal to 1 in the interval \( (0, \frac{\pi}{2}) \), so \( \alpha = \sec^2 x + \csc^2 x - 2 \) must be greater than or equal to 2: \[ \alpha \geq 2 \] Given that \( \alpha \) can grow infinitely as \( x \) approaches 0 or \( \frac{\pi}{2} \) (where \( \tan x \) or \( \cot x \) blow up), \( \alpha \) indeed lies in the interval \( [2, \infty) \).