# Answer to Question #44669 in Trigonometry for ANURAG

Question #44669

Prove that in a triangle with angles A, B and C; and sides of length a, b and c that:

1/[(a-b)(a-c)]*tan(A/2) + 1/[(b-c)(b-a)]*tan(B/2) + 1/[(c-b)(c-a)]*tan(C/2) =(Area of triangle)^(-1)

{NOTE: 'a' is the length of side opposite to angle A, likewise 'b' is the length of side opposite to angle B and similarly 'c' is the length of side opposite to angle C.}

1/[(a-b)(a-c)]*tan(A/2) + 1/[(b-c)(b-a)]*tan(B/2) + 1/[(c-b)(c-a)]*tan(C/2) =(Area of triangle)^(-1)

{NOTE: 'a' is the length of side opposite to angle A, likewise 'b' is the length of side opposite to angle B and similarly 'c' is the length of side opposite to angle C.}

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