# Prove that ( BC - AC )/( BC + AC ) = tan ( A - B )/2/tan ( A + B )/2

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To prove ( BC - AC )/( BC + AC ) = tan ( A - B )/2/tan ( A + B )/2

If ABC is a triangle, then BC = a, AC = b and AB = c.

The sine rule of the triangle is:

a = 2RsinA, b = 2RsinB and C = 2R sinC, where R is a constant (or radius of circumcircle of the triangle.

So (a-b)/(a+b) = (2RsinA-2RsinB)/(2RsinA+2RsinB) = (sinA-sinB)/(sinA+sinB)

(a-b)/(a+b) = (sinA-sinB)/(sinA+sinB)

(a-b)/(a+b)= {2cos(A+B)/2 *sin (A-B)/2}/ {2sin (A+B)/2 * cos(A-B)/2}

(a-b)/(a+b) = {[sin (A-B)/2]/ [cos(A-B)/2]}/ { [sin (A+B)/2]/[cos(A+B)]}

(a-b)/(a+b) = {tan (a-B)/2}/{tan(A+B)/2.}

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]**