In any triangle ABC, Napier's Analogy states that tan((B-C)/2) =

Type: MCQ • Marks: 1

A. (a / (b+c)) * tan(A/2)

B. ((b - c) / (b + c)) * cot(A / 2)

C. sin((B-C)/2) / cos((B+C)/2)

D. ((b + c) / (b - c)) * tan(A / 2)

Explanation

Napier's Analogy relates half-angle differences: tan((B-C)/2) = [(b - c)/(b + c)] cot(A/2).

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