How do you verify the following identity?

$\frac{\cos \left(x\right) - \cos \left(y\right)}{\sin \left(x\right) + \sin \left(y\right)} + \frac{\sin \left(x\right) - \sin \left(y\right)}{\cos \left(x\right) + \cos \left(y\right)} = 0$

Jul 4, 2016

See below.

Explanation:

A helpful starting point when verifying identities involving fractions is to add the fractions. That's what we'll start with here:
$\frac{\cos x - \cos y}{\sin x + \sin y} + \frac{\sin x - \sin y}{\cos x + \cos y} = 0$

$\frac{\left(\cos x - \cos y\right) \left(\cos x + \cos y\right)}{\left(\sin x + \sin y\right) \left(\cos x + \cos y\right)} + \frac{\left(\sin x - \sin y\right) \left(\sin x + \sin y\right)}{\left(\cos x + \cos y\right) \left(\sin x + \sin y\right)} = 0$

$\frac{\left({\cos}^{2} x - {\cos}^{2} y\right) + \left({\sin}^{2} x - {\sin}^{2} y\right)}{\left(\sin x + \sin y\right) \left(\cos x + \cos y\right)} = 0$

As you can see, the binomials in the numerator multiplied very easily because they were differences of squares. Now we rearrange the terms a bit:
$\frac{{\cos}^{2} x - {\cos}^{2} y + {\sin}^{2} x - {\sin}^{2} y}{\left(\sin x + \sin y\right) \left(\cos x + \cos y\right)} = 0$

$\frac{{\cos}^{2} x + {\sin}^{2} x - {\cos}^{2} y - {\sin}^{2} y}{\left(\sin x + \sin y\right) \left(\cos x + \cos y\right)} = 0$

$\frac{{\cos}^{2} x + {\sin}^{2} x - \left({\cos}^{2} y + {\sin}^{2} y\right)}{\left(\sin x + \sin y\right) \left(\cos x + \cos y\right)} = 0$

Recall the Pythagorean Identity ${\sin}^{2} x + {\cos}^{2} x = 1$, or equivalently, ${\sin}^{2} y + {\cos}^{2} y = 1$. Therefore:
$\frac{1 - 1}{\left(\sin x + \sin y\right) \left(\cos x + \cos y\right)} = 0$

$\frac{0}{\left(\sin x + \sin y\right) \left(\cos x + \cos y\right)} = 0$

$0 = 0$

Q.E.D.