# How do you prove  (cos[theta]+tan[theta])/(sin[theta])= sec[theta]+cot[theta]?

May 24, 2016

$\frac{\cos \theta + \sin \frac{\theta}{\cos} \theta}{\sin} \theta = \frac{1}{\cos} \theta + \cos \frac{\theta}{\sin} \theta$

$\frac{\frac{{\cos}^{2} \theta + \sin \theta}{\cos} \theta}{\sin} \theta = \frac{\sin \theta + {\cos}^{2} \theta}{\cos \theta \sin \theta}$

$\frac{{\cos}^{2} \theta + \sin \theta}{\sin \theta \cos \theta} = \frac{\sin \theta + {\cos}^{2} \theta}{\cos \theta \sin \theta}$

Identity proved!!

Hopefully this helps!

May 24, 2016

Here's an alternative way to do it.

Recall that $\sec \theta = \frac{1}{\cos} \theta$, and $\cot \theta = \frac{1}{\tan \theta} = \cos \frac{\theta}{\sin} \theta$.

So, multiplying by $\sin \theta$ gives:

$\frac{\cos \theta + \tan \theta}{\cancel{\left(\sin \theta\right)}} \cdot \cancel{\sin \theta} = \sin \theta \sec \theta + \sin \theta \cot \theta$

$\cos \theta + \tan \theta = \sin \theta \cdot \frac{1}{\cos} \theta + \cancel{\sin \theta} \cdot \cos \frac{\theta}{\cancel{\sin \theta}}$

$\textcolor{b l u e}{\cos \theta + \tan \theta = \tan \theta + \cos \theta}$