How do you prove cos[(pi/2)-x]/sin[(pi/2)-x]=tanx?

Nov 1, 2015

It's because $\cos \left(\frac{\pi}{2} - x\right) = \sin \left(x\right)$, and $\sin \left(\frac{\pi}{2} - x\right) = \cos \left(x\right)$

Explanation:

You need to use two simple trigonometric equality, namely

$\left\{\begin{matrix}\cos \left(\frac{\pi}{2} - x\right) = \sin \left(x\right) \\ \sin \left(\frac{\pi}{2} - x\right) = \cos \left(x\right)\end{matrix}\right.$

Using these two equalities, your expression becomes

$\sin \frac{x}{\cos} \left(x\right)$, which is exactly the definition of $\tan \left(x\right)$