# How do you prove secx - cosx = tanx * sinx?

May 2, 2016

Using the definitions $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ and $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$ along with the identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1 \implies {\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$,
for $\cos \left(x\right) \ne 0$ we have

$\sec \left(x\right) - \cos \left(x\right) = \frac{1}{\cos} \left(x\right) - {\cos}^{2} \frac{x}{\cos} \left(x\right)$

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

$= {\sin}^{2} \frac{x}{\cos} \left(x\right)$

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

$= \tan \left(x\right) \cdot \sin \left(x\right)$