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Question
For triangle ABC, show that:
`tan(("B"+"C")/2) = cot("A"/2)`
Sum
Solution
We know that for a triangle ΔABC
∠A + ∠B + ∠C = 180°
∠B + ∠C = 180° - ∠A
`(∠"B" + ∠"C")/(2) = 90° - (∠"A")/(2)`
`tan(("B"+"C")/(2)) = tan (90° - "A"/(2))`
`tan(("B"+"C")/(2)) = cot ("A"/2)`
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Complimentary Angles for Tangent ( Tan ) and Contangency ( Cot )
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