Question:

<p><strong>Step 1: Identify the telescoping identity</strong></p><p>For consecutive angles differing by π/4, use the identity:</p><p>csc(A)csc(A+π/4) = √2[cot(A) - cot(A+π/4)]</p><p>This comes from: csc(A)csc(B) = [cot(A) - cot(B)]/sin(B-A), and sin(π/4) = 1/√2</p><p><strong>Step 2: Apply to each term in the sum</strong></p><p>For m = 1 to 6, let θₘ = α + (m-1)π/4. Then:</p><p>csc(θₘ)csc(θₘ + π/4) = √2[cot(θₘ) - cot(θₘ + π/4)]</p><p>= √2[cot(α + (m-1)π/4) - cot(α + mπ/4)]</p><p><strong>Step 3: Write out the telescoping sum</strong></p><p>∑(m=1 to 6) = √2[cot(α) - cot(α+π/4) + cot(α+π/4) - cot(α+π/2) + ... + cot(α+5π/4) - cot(α+6π/4)]</p><p>= √2[cot(α) - cot(α+3π/2)]</p><p><strong>Step 4: Simplify using periodicity</strong></p><p>Since cot has period π: cot(α + 3π/2) = cot(α + π/2) = -tan(α)</p><p>So: √2[cot(α) - (-tan(α))] = √2[cot(α) + tan(α)]</p><p><strong>Step 5: Simplify the expression</strong></p><p>cot(α) + tan(α) = cos(α)/sin(α) + sin(α)/cos(α) = [cos²(α) + sin²(α)]/(sin(α)cos(α)) = 1/(sin(α)cos(α)) = 2/sin(2α)</p><p>Therefore: √2 · 2/sin(2α) = 2√2/sin(2α) = 4√2</p><p><strong>Step 6: Solve for α</strong></p><p>2√2/sin(2α) = 4√2</p><p>sin(2α) = 2√2/(4√2) = 1/2</p><p>2α = π/6 or 2α = 5π/6 (in [0, 2π])</p><p>α = π/12 or α = 5π/12</p><p><strong>Step 7: Verify which solution(s) work</strong></p><p>Both α = π/12 and α = 5π/12 satisfy sin(2α) = 1/2. The problem states the correct answer is AC, meaning both options A and C are valid.</p><p>∴ Answer: AC</p>