If cot²θ - 2 cos²θ = 0, where 0°

Question

If cot²θ - 2 cos²θ = 0, where 0° < θ < 90°, then find the value of θ.

Accepted Answer

Given the equation: cot²θ - 2 cos²θ = 0 Step 1: Express cot²θ in terms of sine and cosine. We know cotθ = cosθ / sinθ, so cot²θ = cos²θ / sin²θ. Step 2: Substitute cot²θ: (cos²θ / sin²θ) - 2 cos²θ = 0 Step 3: Multiply through by sin²θ to clear denominator: cos²θ - 2 cos²θ sin²θ = 0 Step 4: Factor out cos²θ: cos²θ (1 - 2 sin²θ) = 0 Step 5: For the product to be zero, either: (i) cos²θ = 0 or (ii) 1 - 2 sin²θ = 0 Step 6: cos²θ = 0 implies cosθ = 0, which means θ = 90°, but θ must be between 0° and 90°, so θ = 90° is not valid as cos90°=0 but cot90° is undefined. Step 7: Solve 1 - 2 sin²θ = 0 => 2 sin²θ = 1 => sin²θ = 1/2 => sinθ = ±1/√2 Since 0° < θ < 90°, sinθ = 1/√2 Step 8: sinθ = 1/√2 corresponds to θ = 45° Therefore, the value of θ is 45°.

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