# How do you prove (csc(-t)-sin(-t))/sin(t) = -cot^2(t)?

May 16, 2015

Remember that $\sin$ is an odd function so $\sin \left(- t\right) = - \sin \left(t\right)$ and that $\cot = \frac{\cos}{\sin}$ and $\csc = \frac{1}{\sin}$ so you get:

$\frac{\frac{1}{\sin} \left(- t\right) - \sin \left(- t\right)}{\sin} \left(t\right) = - {\cos}^{2} \frac{t}{\sin} ^ 2 \left(t\right)$
$\frac{- \frac{1}{\sin} \left(t\right) + \sin \left(t\right)}{\sin} \left(t\right) = - {\cos}^{2} \frac{t}{\sin} ^ 2 \left(t\right)$
$\frac{- 1 + {\sin}^{2} \left(t\right)}{\cancel{{\sin}^{2} \left(t\right)}} = - {\cos}^{2} \frac{t}{\cancel{{\sin}^{2} \left(t\right)}}$
$- {\cos}^{2} \left(t\right) = - {\cos}^{2} \left(t\right)$

Where I used the fact that: ${\sin}^{2} \left(t\right) + {\cos}^{2} \left(t\right) = 1$
and: ${\cos}^{2} \left(t\right) = 1 - {\sin}^{2} \left(t\right)$
so $- {\cos}^{2} \left(t\right) = - 1 + {\sin}^{2} \left(t\right)$