In a triangle ABC, sin A = cos B + cos C then what is tan(B/2) + cot(B/2) equal to?
A.1
B.√2
C.√3
D.2✓ Correct
Explanation
sin A = cos B + cos C. Using A = π - (B+C): sin(B+C) = cos B + cos C. 2 sin((B+C)/2) cos((B+C)/2) = 2 cos((B+C)/2) cos((B-C)/2). So sin((B+C)/2) = cos((B-C)/2). This gives cos((π-A)/2 - ...) and leads to B = π/2. Then tan(B/2) + cot(B/2) = tan(π/4) + cot(π/4) = 1 + 1 = 2.
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