Please prove the following trigonometric identity. Please give step by step answer.?

Question

(sin A + sin B)/(sin A -sin B) =[tan (A+B)/2]/[ tan (A+B)/2]

To all respondents - I am so sorry. I made a big typo/error. The right hand side denominator is tan (A-B)/2.

Answer 1

Assuming that it should actually be:

(sin A + sin B)/(sin A -sin B) =[tan (A+B)/2]/[ tan (A-B)/2]

Then it requires knowing some of the sum-to-product identities, namely:

sin A + sin B = 2sin[(A + B)/2]cos[(A - B)/2]
sin A - sin B = 2cos[(A + B)/2]sin[(A - B)/2]

(sin A + sin B)/(sin A -sin B) = (2sin[(A + B)/2]cos[(A - B)/2])/
(2cos[(A + B)/2]sin[(A - B)/2])


(sin A + sin B)/(sin A -sin B) = (sin[(A + B)/2]cos[(A - B)/2])/
(cos[(A + B)/2]sin[(A - B)/2])

(sin A + sin B)/(sin A -sin B) = tan[(A + B)/2]cot[(A - B)/2]

(sin A + sin B)/(sin A -sin B) = (tan[(A + B)/2])/(tan[(A - B)/2])

Answer 2

because a million-cot A = (tan A-a million)/tan A so convert so as that both have similar denominator 1st time period tan A /(a million- cot A) = tan A/(a million- a million/tan A) = tan^2 A/(tan A-a million) upload the 2d time period to get LHS = tan^2 A/(tan A -a million) - cot A.(tan A-a million) = (tan^2A-a million/tan A)(tan A-a million) = (tan^3A -a million)/(tan A(tan A-a million) = (tan^2A + tan A+ a million)/tan A(utilizing (a^3-b^3)/(a-b) = a million+(tan^2 A+a million)/tan A = a million+ sec^2A/tan A (isolating a million by way of the indisputable fact that is on RHS) = a million+ a million/cos^2A. cos A/sin A = a million +a million/(cos A sin A) = a million + sec A cosec A = RHS

Answer 3

I think the angles must belong to a triangle.
Then A+B+C=180
A+B=180-C
(A+B)/2= 90-C/2
tan(A+B)/2=tan(90-C/2)

On the right hand side one + sign
must me minus, isn't it?

Answer 4

(sin A + sin B) = 2sin[(A+B)/2] * cos[(A-B )/2]
(sin A - sin B) = 2sin[(A-B)/2] * cos[(A+B )/2]
thus
(sin A + sin B)/(sin A -sin B) = sin[(A+B)/2] / cos[(A+B )/2] * cos[(A-B )/2] / sin[(A-B)/2] = [tan (A+B)/2]/[ tan (A+B)/2]

Answer 5

are you sure you posted that right?

Because the right hand side of what you posted simplifies to 1, and the left side does not.

Answer 6

RHS=1, but LHS is not necessarily 1.

Something is wrong here.

Answer 7

sinA+sinB
=2sin(A+B)/2cos(A-B)/2
sinA-sinB
=2cos(A+B)/2sin(A-B)/2
so sin(A+B)/sin(A-B)
=2sin(A+B)/2cos(A-B)/
2/2cos(A+B)/2sin(A-B)/2
=[2sin(A+B)/2/cos(A+B)/2]/
[2sin(A-B)/2cos(A-B)/2]
=[tan(A+B)/2]/[tan(A-B)/2]