# In a triangle if A,B,C are angles and a,b.c are the corresponding opposite sides and if (b+c) = 3a, Calculate cot(B/2) cot(C/2) = 2?

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- Pramod KumarLv 77 months ago
Since (b+c) = 3a , we have sin B+sin C = 3 sin A

=> 2 sin [ (B+C)/2 ] . cos [ (B-C)/2 ] = 6 sin A/2 . cos A/2

=> cos [ (B-C)/2 ] = 3 sin A/2 .... .. since [ sin (B+C)/2 ] = cos A/2

=> cos [ (B-C)/2 ] = 3 cos [ (B+C)/2 ]

Now expand cos expressions -

=> cos (B/2) . cos (C/2) + sin (B/2) .sin (C/2) = 3

cos(B/2).cos (C/2) - 3 sin (B/2).sin (C/2)

Divide both the sides by (sin B/2 . sin C/2)

=> cot (B/2) . cot (C/2) + 1 = 3 cot (B/2) cot (C/2) - 3

=> cot (B/2) . cot (C/2) = 2

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