Question

For ΔABC, prove that tan(A+C/2) = cot*b/2

Answer 1

(a-b)/(a+b)=(2R sin A - 2R sin B)/(2R sin A + 2R sin B)
=(sin A - sin b) / ( sin A + sin B)
= 2 cos (A+B /2)sin (A - B / 2) / 2sin(A+B /2)cos (A- B / 2)
= cot (A+B / 2)tan(A-B / 2)
=tan (C/2) tan (A - B /2) [A+B+C=pi

(a-b)/(a+b)cot(C/2)=tan(A-B/2)

Answer 2

In ΔABC ,

∠A + ∠B + ∠C = 180° [ Angle sum property ]

⇒ ∠A +∠C = 180° - ∠B

Divide each term with 2 , we get

⇒ ∠A/2 + ∠C/2 = 180°/2 - ∠B/2

⇒ ( ∠A + ∠C )/2 = 90° - ∠B/2

⇒ Tan ( ∠A + ∠C )/2 = Tan ( 90° - ∠B/ 2 )

⇒ Tan ( ∠A + ∠C )/2 = Cot ∠B/2

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