Question

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.}

Answer 1

Answer on Question #44669 – Math - Trigonometry

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

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

{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.}

Solution.

According to the Law of Cosines:

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$

As we know

$$\cos A = 2 \cos^2 \frac{A}{2} - 1 \rightarrow \cos^2 \frac{A}{2} = \frac{1 + \cos A}{2} = \frac{b^2 + c^2 - a^2 + 2bc}{4bc} = \frac{(b + c - a)(b + c + a)}{4bc} = \frac{s(s - a)}{bc}, \quad \text{where}$$ $$s = \frac{a + b + c}{2}.$$ $$\cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}, \quad \sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}},$$ $$\tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} = \frac{P}{s(s - a)}$$

Where $P = \sqrt{s(s - a)(s - b)(s - c)}$ - area of the triangle (Heron’s formula).

Similarly: $\tan \frac{B}{2} = \frac{P}{s(s - b)}$, $\tan \frac{C}{2} = \frac{P}{s(s - c)}$.

So, $\frac{\tan_2^A}{(a - b)(a - c)} + \frac{\tan_2^B}{(b - c)(b - a)} + \frac{\tan_c^C}{(c - b)(c - a)} =$

$$= \frac{P}{s} \left[ \frac{1}{(a - b)(a - c)(s - a)} + \frac{1}{(b - a)(b - c)(s - b)} + \frac{1}{(c - a)(c - b)(s - c)} \right] =$$ $$\begin{array}{l} = \frac {P}{s} * \frac {- (b - c) (s - b) (s - c) - (c - a) (s - a) (s - c) - (a - b) (s - a) (s - b)}{(a - b) (b - c) (c - a) (s - a) (s - b) (s - c)} = \\ = - \frac {P}{s (s - a) (s - b) (s - c)} * \\ * \frac {(b - c) (a - b + c) (a + b - c) + (c - a) (b + c - a) (a + b - c) + (a - b) (b + c - a) (a - b + c)}{4 (a - b) (b - c) (c - a)} \\ = - \frac {P}{P ^ {2}} * \frac {4 a c ^ {2} + 4 a ^ {2} b - 4 a ^ {2} c + 4 b ^ {2} c - 4 b c ^ {2} + 4 a b ^ {2}}{4 (a - b) (b - c) (c - a)} = \frac {1}{P} * \frac {4 (a - b) (b - c) (c - a)}{4 (a - b) (b - c) (c - a)} = \\ = \frac {1}{P}. \end{array}$$

Q.E.D.

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