Let S={θ ∈(0, π/2): Σₘ₌₁⁹ sec (θ+(m-1) π/6) sec (θ+(m π)/6)=-8/√3}. Then

Solution

Maths sir · 5 steps

Turn each sec-times-sec product into a difference of tangents, and the whole sum telescopes.

1. Make it telescope. In each term the two angles are A and B = A + π/6, so sin(B-A) = 1/2. Then sec A sec B = sin(B-A)/(1/2 cos A cos B) = 2(tan B - tan A).

   Σₘ₌₁⁹ = 2[tan(θ+3π/2)-tan θ] = -2(cot θ+tan θ)

2. Collapse the sum. All the middle tangents cancel. Only the first angle θ and the last angle θ + 9π/6 = θ + 3π/2 remain, and tan(θ + 3π/2) = -cot θ.
3. Solve for theta. tan θ + cot θ = 1/(sin θ cos θ) = 2/(sin 2θ), so -4/(sin 2θ) = -8/√3 and sin 2θ = √3/2. With 2θ ∈ (0, π), 2θ = π/3 or 2π/3, so θ = π/6 or π/3.
4. A caution. Strictly, both values make one secant in the original sum equal to sec π/2, which is undefined. The official key ignores this and takes the roots of the reduced equation, so we do the same.
5. So. S = {π/6, π/3} and the sum of its elements is π/6 + π/3 = π/2, which is option (c).

The idea to keep. When consecutive angles differ by a fixed d, use sec A sec B = (tan B-tan A)/(sin d). The sum then telescopes, leaving only the first and last angles.

Answer from the NTA final answer key. Solution by PrepWiser.