For an acute angle θ, suppose tan^2 θ + cot^2 θ = 2. Find the value of 2 sec θ cosec θ. Choose the correct value.

Introduction / Context:
This question tests connecting algebraic conditions on tan and cot to a specific trig expression involving sec and cosec. The key observation is that for real numbers, t^2 + 1/t^2 is at least 2, and it equals 2 only in a special case. Once that special case is found, the angle becomes a standard angle and the remaining expression is easy to compute.

Given Data / Assumptions:

• θ is acute (0° < θ < 90°) • tan^2 θ + cot^2 θ = 2 • Required: 2 sec θ cosec θ

Concept / Approach:
Let t = tan θ (> 0 for acute θ). Then cot θ = 1/t. So the condition becomes: t^2 + 1/t^2 = 2. But (t − 1/t)^2 = t^2 − 2 + 1/t^2. If t^2 + 1/t^2 = 2, then (t − 1/t)^2 = 0, meaning t = 1. That implies θ = 45°. Then compute sec and cosec at 45° and plug into 2 sec θ cosec θ.

Step-by-Step Solution:
1) Let t = tan θ, so cot θ = 1/t 2) Given: t^2 + 1/t^2 = 2 3) Compute (t − 1/t)^2: (t − 1/t)^2 = t^2 − 2 + 1/t^2 4) Substitute t^2 + 1/t^2 = 2: (t − 1/t)^2 = 2 − 2 = 0 5) Therefore t − 1/t = 0 → t = 1 6) tan θ = 1 and θ acute implies θ = 45° 7) At 45°: sec θ = 1/cos θ = √2 and cosec θ = 1/sin θ = √2 8) So 2 sec θ cosec θ = 2 * √2 * √2 = 2 * 2 = 4

Verification / Alternative check:
If θ = 45°, tan^2 θ = 1 and cot^2 θ = 1, so the condition tan^2 θ + cot^2 θ = 2 holds exactly. Then the computed value 4 is consistent.

Why Other Options Are Wrong:
• 0 or 1: too small; the expression equals 4 at θ = 45°. • 2 or 3: would require different θ, but the condition forces tan θ = 1 uniquely for acute θ.

Common Pitfalls:
• Assuming many angles satisfy tan^2 + cot^2 = 2 (actually only tan θ = 1 works for acute θ). • Forgetting that sec 45° = √2 and cosec 45° = √2.

Final Answer:
4