# How do you prove Sec(x) - cos(x) = sin(x) * tan(x)?

Jul 10, 2016

Knowing that $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ and $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$, rewriting the equation yields

$\frac{1}{\cos} \left(x\right) - \cos \frac{x}{1} = \sin \left(x\right) \cdot \left(\sin \frac{x}{\cos} \left(x\right)\right)$

Rewriting the left fraction by using the property

$\frac{a}{b} - \frac{c}{d} = \frac{a d - b c}{b d}$

and simplifying the right side of the equation yields

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

Note that in this case, we can make use of the identity

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$, since $1 - {\cos}^{2} \left(x\right) = {\sin}^{2} \left(x\right)$, giving us

${\sin}^{2} \frac{x}{\cos} \left(x\right) = {\sin}^{2} \frac{x}{\cos} \left(x\right)$, which is true, therefore we have proven that

$\sec \left(x\right) - \cos \left(x\right) = \sin \left(x\right) \tan \left(x\right)$