Need to prove the identity Sin(x+y)/sin(x-y)=(tanx+tany)/(tanx-tany)?

Question

step by step please

Answer 1

Turn the right hand side into sin's and cos's.

(tan x + tan y)/(tan x - tan y) = (sin x/cos x + sin y/cos y)/(sin x/cos x - sin y/cos y)
Multiply top and bottom by cos x cos y to get rid of the fractions:
= (sin x cos y + sin y cos x)/(sin x cos y - sin y cos x).
And now you recognise that the top is the identity for sin(x+y) and the bottom is the identity for sin(x-y).

Answer 2

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Answer 3

(tan x + tan y)
----------------
(tan x - tan y)

(sin x/cos x + sin y/cos y)
= ---------------------------
(sin x/cos x - sin y/cos y)

Numerator is:
sin x/cos x + sin y/cos y
= (sin x cos y + sin y cos x)/(cos x cos y)

and denominator is:
(sin x/cos x - sin y/cos y)
= (sin x cos y - sin y cos x)/(cos x cos y)

Plug these in to get:

(sin x cos y + sin y cos x)/(cos x cos y)
= -----------------------------------------
(sin x cos y - sin y cos x)/(cos x cos y)

(sin x cos y + sin y cos x)
= ---------------------------
(sin x cos y - sin y cos x)

= sin(x + y) / sin(x - y)

So RHS = LHS

Proved