# Question #5cf65

Nov 27, 2016

Currently, as is, the identity is false. If we change $\sin x - \cos x$ in the denominator on the left to $\sin x + \cos x$, it will be true though (thanks to Mrs. R. for pointing this out).

We know that $\tan \theta = \sin \frac{\theta}{\cos} \theta$, $\cot \theta = \frac{1}{\tan} \theta = \frac{1}{\sin \frac{\theta}{\cos} \theta} = \cos \frac{\theta}{\sin} \theta$, $\sec \theta = \frac{1}{\cos} \theta$ and $\csc \theta = \frac{1}{\sin} \theta$. So:

$\frac{\cos \frac{x}{\sin} x - \sin \frac{x}{\cos} x}{\sin x + \cos x} = \frac{1}{\sin} x - \frac{1}{\cos} x$

$\frac{\frac{{\cos}^{2} x - {\sin}^{2} x}{\sin x \cos x}}{\sin x + \cos x} = \frac{\cos x - \sin x}{\cos x \sin x}$

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

$\frac{\cos x - \sin x}{\sin x \cos x} = \frac{\cos x - \sin x}{\cos x \sin x}$

$L H S = R H S$

Identity proved!

Hopefully this helps!