How do you prove the identity: (sin(x+y))/(sin(x-y)) = (1+cot x tan y)/(1-cot x tan y)?

Answer 1

First expand the left hand side to what the sum and difference formulas are:

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

Then take each term on the right hand side and divide it by sin(x)cos(y)


You then get [ sin(x)cos(y) / sin(x)cos(y) ] + [ cos(x)sin(y) / sin(x)cos(y) ] as your numerator.

AND [ sin(x)cos(y) / sin(x)cos(y) ] - [ cos(x)sin(y) / sin(x)cos(y) ] as your denominator.

These simplify to : Numerator : 1 + cot(x)tan(y) and Denominator: 1 - cot(x)tan(y).

Hence

1 + cot(x)tan(y)
-----------------------
1 - cot(x)tan(y)

Answer 2

Hope you know the expansions for sin(x + y) etc. Using these,

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

Now divide each term in the numerator and the denominator by
sin x cos y. The first term in both num and denom is then 1,
and the second term in each is (cos x sin y)/(sin x cos y).

Now cos x / sin x = tan x, and sin y / cos y = tan y,
and so you have the result.

Answer 3

Are you taking PreCalculus. The book has a formula about sin(A+B) and sin(A-B). Find that out. There should be formula for cot and tan.

Answer 4

LHS = (sin (x+y))/(sin (x-y)
= (sinx.cosy + cosx.siny)/(sinx.cosy - cosx.siny)

= (sinx.cosy/sinx.cosy + cosx.siny/sinx.cosy)/(sinx.cosy/sinx.cosy - cosx.siny/sinx.cosy)
(on dividing numerator and denominator by sinx.cosy

= [1 + (cosx/sinx)(siny/cosy)]/[1 - (cosx/sinx)(siny/cosy)]
= (1 + cot x tan y)/(1 - cot x tan y)
= RHS .... QED

Answer 5

Graphing calculator.