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Question
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
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Solution
We know that for a triangle ΔABC
∠A + ∠B + ∠C = 180°
∠B + ∠C = 180° – ∠A
`=> (angle B + angle C)/2 = 90^circ - (angle A)/2`
`=> tan ((B + C)/2) = tan (90^circ - A/2)`
= `cot (A/2)`
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