# `(cos(x) - cos(y))/(sin(x) + sin(y)) + (sin(x) - sin(y))/(cos(x) + cos(y)) = 0` Verfiy the identity.

*print*Print*list*Cite

### 2 Answers

We'll use the formula `(a-b)*(a+b) = a^2-b^2.`

Multiply equation by `(sin(x)+sin(y))*(cos(x)+cos(y))` (the product of the denominators):

`(cos(x)+cos(y))*(cos(x)-cos(y)) +(sin(x)+sin(y))*(sin(x)-sin(y)) = 0,`

`(cos^2(x)-cos^2(y)) + (sin^2(x)-sin^2(y)) = 0,`

`(cos^2(x)+sin^2(x)) - (cos^2(y)+sin^2(y)) = 0,`

1-1 = 0, QED.

### Hide Replies ▲

By multiplying both sides of the equation by the product of the denominators, you are assuming the equation is true. But that is what you are trying to prove.

Better is to rewrite the left side as a single fraction and note that the numerator is 0 (1-1) and the denominator is non-zero thus establishing the identity.

By cross multiplication ,we get

`((cosx-coy)(cosx+cosy))+((sinx-siny)(sinx+siny))/((sinx+siny)(cosx+cosy))=`

`(cos^2x-cos^2y+sin^2x-sin^2y)/((sinx+siny)(cosx+cosy))=((cos^2x+sin^2x)-(cos^2y+sin^2y))/((sinx+siny)(cosx+cosy))`

`=(1-1)/((sinx+siny)(cosx+cosy))=0.`