# How do you verify the identity [(cosx cotx)/1-sin x]-1=cscx?

Jun 12, 2015

Is there a typo here? These are not equal.

http://www.wolframalpha.com/input/?i=Is+[cosxcotx+-+sinx]+-+1+%3D+cscx

With $\cot x = \cos \frac{x}{\sin} x$ and $\csc x = \frac{1}{\sin} x$:

$\cos x \cot x - \sin x - 1 = \csc x$

${\cos}^{2} \frac{x}{\sin} x - \sin x - 1 = \csc x$

${\cos}^{2} \frac{x}{\sin} x - \sin x - 1 = \frac{1}{\sin} x$

${\cos}^{2} x - {\sin}^{2} x - \sin x = 1$

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

${\cos}^{2} x \ne \left[\sin x + 2 {\sin}^{2} x\right] + {\cos}^{2} x$