We notice that BC is the opposite side of the angle A and
AC is the opposite side of the angle B.
This relation
between the tangents of 2 angles of a triangle and the opposite sides is called the law
of tangents.
We'll note the length of the side BC = a and
the length of the side AC = b.
We'll re-write the given
relation:
(a-b)/(a+b) = [tan (A - B)/2]/[tan (A + B)/2]
(1)
We'll apply the law of sines to write the lengths of
the sides a and b.
a/sin A = b/sin B =
2R
a = 2R*sin A (2)
b = 2R*sin
B (3)
We'll substitute (2) and (3) in
(1).
(2R*sin A - 2R*sin B)/(2R*sin A + 2R*sin B) = [tan (A
- B)/2]/[tan (A + B)/2]
We'll factorize by 2R to the left
side and we'll simplify:
(sin A - sin B)/(sin A + sin B) =
[tan (A - B)/2]/[tan (A + B)/2]
We'll transform in product
the sum and the difference of sines:
sin A - sin B = 2*cos
[(A+B)/2]*sin [(A-B)/2] (4)
sin A + sin B = 2*sin
[(A+B)/2]*cos [(A-B)/2] (5)
We'll divide (4) by
(5):
2*cos [(A+B)/2]*sin [(A-B)/2]/2*sin [(A+B)/2]*cos
[(A-B)/2] = [tan (A - B)/2]/[tan (A + B)/2]
We'll simplify
and we'll write each quotient:
cos [(A+B)/2]/sin [(A+B)/2]
= cot [(A+B)/2]
cot [(A+B)/2] = 1/ tan [(A+B)/2]
(6)
sin [(A-B)/2]/cos [(A-B)/2] = tan [(A-B)/2]
(7)
We'll multiply (6) by
(7):
{1/ tan [(A+B)/2]}*tan [(A-B)/2]=[tan (A
- B)/2]/[tan (A + B)/2]
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