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Question:

If cosecθ = p+q/p-q (p≠q≠0), then cot(π/4+θ/2) is equal to:

√pq

pq

√qp

√pq

Solution:

cosecθ = p+q/p-q, sinθ = p-q/p+q
cot(θ/2+π/4) = cot(π/4)cot(θ/2) -1/cot(π/4)+cot(θ/2) = cot(θ/2)-1/cot(θ/2)+1.
∴sinθ = p-q/p+q ∴cosθ = √1-sin²θ = √1-(p-q/p+q)² = 2√pq/p+q
cosθ = 2cos²(θ/2)-1 ∴cos²(θ/2) = 1/2(2√pq/p+q+1) = 1/2(√p+√q)²/p+q
sin²(θ/2) = 1/2(1-2√pq/p+q) = 1/2(√p-√q)²/p+q
cot²θ = (√p+√q)²/(√p-√q)² ∴cotθ = (√p+√q)/(√p-√q)
cot(θ/2) = √(1+cosθ)/(1-cosθ) = √(1+2√pq/p+q)/(1-2√pq/p+q) = (√p+√q)/(√p-√q)
cot(θ/2) + 1 = (√p+√q)/(√p-√q) + 1 = 2√p/(√p-√q)
cot(θ/2) -1 = (√p+√q)/(√p-√q) - 1 = 2√q/(√p-√q)
cot(θ/2+π/4) = cot(θ/2)-1/cot(θ/2)+1 = (2√q/(√p-√q))/(2√p/(√p-√q)) = √q/√p = √qp