Breaking a Triangular gate #2
In a triangle ABC , is equal to :
Note: Sides are opposite of angles
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1 solution
a, b, c are the opposite of angles A, B, C in a triangle ABC.
tan (A/2) x tan (B/2) = ((1 - cos A) / sin A) x (sin B / (1 + cos B)) = (sin B - (sin B)(cos A)) / (sin A + (sin A)(cos B)) = (sin B - (1/2)(sin (B+A) + sin (B-A)) / (sin A + (1/2)(sin (A+B) + sin (A-B)) = (sin B - (1/2)(sin (180-C) + sin -(B-A)) / (sin (180-C) + (1/2)(sin (A+B) + sin (A-B)) = (sin B - (1/2)(sin C - sin (A-B)) / (sin A + (1/2)(sin C + sin (A-B)) = (sin B - (1/2)(sin C) + (1/2)(sin (A-B)) / (sin A + (1/2)(sin C) + (1/2)(sin (A-B))
sin (A-B) can be replaced with sin (180 - (A-B)) = sin (180-A+B) = sin (B+C+B) = sin (2B+C). sin (A-B) is the same with sin (2B+C).
Based on the rule of sinus : a / sin A = b / sin B = c / sin C, we can assume: sin A is equivalent to a; sin B is equivalent to b; sin C is equivalent to c, and so on. Thus, the equation will be: 1. (sin B - (1/2)(sin C) + (1/2)(sin (A-B)) / (sin A + (1/2)(sin C) + (1/2)(sin (A-B)) = (b - 1/2 (c) + 1/2 (a-b)) / (a + 1/2 (c) + 1/2 (a-b)) = (1/2 a + 1/2 b - 1/2 c) / (3/2 a - 1/2 b + 1/2 c) = (a + b - c) / (3a - b + c) It can be manipulated by the rule of sinus: (a + b - c) / (3a - b + c) is equivalent to sin (A + B - C) / sin (3A - B + C) <=> sin (A+B-C) / sin (3A-B+C) = sin (A+B-C) / sin ((2A+C)+(A-B)) = sin (A+B-C) / sin ((2A+C)+(2B+C)) = sin (A+B-C) / sin (2A+2B+2C)
A, B, C are the angles of ABC, with A+B+C = 180 degrees. sin (2A+2B+2C) = sin 2(A+B+C) = sin 2.180 = sin 360; and sin 360 = sin 180 = sin 0 = 0. it means that sin (2A+2B+2C) is the same with sin (A+B+C). The coefficient 2 doesn't affect the value, so: = sin (A+B-C) / sin (2A+2B+2C) = sin (A+B-C) / sin (A+B+C) Based the rule of sinus, sin (A+B-C) / sin (A+B+C) is equivalent to (a+b-c) / (a+b+c).
=> tan (A/2). tan (B/2) = (a + b - c)/(a + b + c)