# How do you verify sin x + cos x + tan x sin x = sec x + cos x tan x?

Jul 22, 2015

Start with the identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$
and gradually work towards the given equation

#### Explanation:

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

divide everything by $\cos \left(x\right)$ ...we'll worry about the case when $\textcolor{red}{\cos \left(x\right) = 0}$ later
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{b l u e}{{\sin}^{2} \frac{x}{\cos} \left(x\right) + \cos \left(x\right) = \frac{1}{\cos} \left(x\right)}$

since $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$ and $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{b l u e}{\sin \left(x\right) \tan \left(x\right) + \cos \left(x\right) = \sec \left(x\right)}$

rearrange to look closer to our target equation:

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{b l u e}{\cos \left(x\right) + \tan \left(x\right) \sin \left(x\right) = \sec \left(x\right)}$

looks like we need to add $\sin \left(x\right)$ to make the left side look like the target

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{b l u e}{\sin \left(x\right) + \cos \left(x\right) + \tan \left(x\right) \sin \left(x\right) = \sec \left(x\right) + \sin \left(x\right)}$

since $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$
therefore #sin(x) = cos(x)tan(x)

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{b l u e}{\sin \left(x\right) + \cos \left(x\right) + \tan \left(x\right) \sin \left(x\right) = \sec \left(x\right) + \cos \left(x\right) \tan \left(x\right)}$

We have now proven the target equation except for the case $\textcolor{red}{\cos \left(x\right) = 0}$

If $\cos \left(x\right) = 0$ then $\tan \left(x\right)$ is undefined
and since $\tan \left(x\right)$ appears on both sides
$\textcolor{w h i t e}{\text{XXXX}}$we will pretend that $\text{undefined" = "undefined}$
$\textcolor{w h i t e}{\text{XXXX}}$is adequate to cover this case.