TrigonometryFind if (BC-AC)/(BC+AC) = tan(A-B)/2/tan(A+B)/2

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We notice that BC is the opposite side to angle A and AC is the opposite side to the angle B.

We'll use the 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 reduce:

(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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