How do you verify the identity (cosx-cosy)/(sinx+siny)+(sinx-siny)/(cosx+cosy)=0?

Jan 4, 2017

Put on a common denominator.

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

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

Use the pythagorean identity ${\sin}^{2} x + {\cos}^{2} x = 1$:

$\frac{1 - {\cos}^{2} y - {\sin}^{2} y}{\left(\sin x + \sin y\right) \left(\cos x + \cos y\right)} = 0$

Use the pythagorean identity mentioned above again, except this time in the form ${\sin}^{2} x = 1 - {\cos}^{2} x$.

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

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

$0 = 0$

$L H S = R H S$

Identity proved!

Hopefully this helps!

Jan 4, 2017

See below.

Explanation:

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