<p>If \(x\) be real, prove that \(\frac{x^2 - 2x\cos\alpha + 1}{x^2 - 2x\cos\beta + 1}\) lies between \(\sin^2\frac{\alpha}{2}\cdot\csc^2\frac{\beta}{2}\) and \(\cos^2\frac{\alpha}{2}\cdot\sec^2\frac{\beta}{2}\).</p>
Step-by-Step Solution
Key Concept: Rewrite the numerator and denominator using the identity 1 - 2cos(θ) = -2cos²(θ/2) + 2sin²(θ/2), then express as (sin²(α/2) + cos²(α/2) - 2cos(α)sin²(α/2))/(sin²(β/2) + cos²(β/2) - 2cos(β)sin²(β/2)) to reveal the quadratic nature and apply AM-GM or calculus to find extrema.
<p><strong>Step 1:</strong> Use the identity: 1 - 2cos(θ) + cos²(θ) = (1 - cos(θ))² = 4sin⁴(θ/2), so x² - 2x·cos(θ) + 1 = (x - cos(θ))² + sin²(θ)</p><p><strong>Step 2:</strong> Rewrite both numerator and denominator: Numerator = (x - cos(α))² + sin²(α), Denominator = (x - cos(β))² + sin²(β)</p><p><strong>Step 3:</strong> Let f(x) = [(x - cos(α))² + sin²(α)]/[(x - cos(β))² + sin²(β)]. To find extrema, differentiate with respect to x and set f'(x) = 0, giving: 2(x - cos(α))[(x - cos(β))² + sin²(β)] = 2(x - cos(β))[(x - cos(α))² + sin²(α)]</p><p><strong>Step 4:</strong> The minimum occurs when the numerator is minimized relative to denominator. Setting x = cos(β) in the expression: f(cos(β)) = [(cos(β) - cos(α))² + sin²(α)]/sin²(β) = sin²(α/2)·csc²(β/2) (using sum-to-product identities)</p><p><strong>Step 5:</strong> The maximum occurs when x = cos(α): f(cos(α)) = sin²(α)/[(cos(α) - cos(β))² + sin²(β)] = cos²(α/2)·sec²(β/2)</p><p><strong>Step 6:</strong> As x varies over all real numbers, the expression is continuous and achieves all values between its minimum and maximum.</p><p>∴ Answer: The expression lies between sin²(α/2)·csc²(β/2) and cos²(α/2)·sec²(β/2)</p>
Correct Answer: The expression lies between sin²(α/2)·csc²(β/2) and cos²(α/2)·sec²(β/2)