Question 15

We use 1/(sin k° sin(k+1)°) = (cot k° − cot(k+1)°)/sin 1°. So α = (1/sin1°)[(cot60 − cot61) + (cot62 − cot63) + ... + (cot118 − cot119)]. Pairing terms: cot(120−x) = −cot(60+x)... Let's check: terms with k = 60, 62, ..., 118 (30 terms). Sum S = Σ (cot k − cot(k+1)) for k = 60, 62, ..., 118. Note cot(180−θ) = −cot θ. Pair k = 60 with k = 118: cot60 − cot61 + cot118 − cot119 = cot60 − cot61 − cot62 + cot61 = cot60 − cot62. Similarly k=62 with k=116: cot62 − cot63 + cot116 − cot117 = cot62 − cot63 − cot64 + cot63 = cot62 − cot64. Continuing, sum telescopes: for pairs k and 178−k, giving cot k − cot(k+2). Pairs: (60,118),(62,116),...,(88,90): last pair k=88 with 90: cot88 − cot89 + cot90 − cot91 = cot88 − cot89 + 0 + cot89 = cot88. Hmm, consistent with cot88 − cot90 = cot88. So total S = cot60 − cot90 = cot60 = 1/√3. Thus α = 1/(√3 sin1°), so cosec1°/α = √3, squared = 3.

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