# How do you prove sec x + sin x = tan x sin x?

Apr 26, 2016

This is true only for $x = - \setminus \frac{\pi}{4}$ plus multiples of $\setminus \pi$.

#### Explanation:

Since $\setminus \sec \left(x\right) = \frac{1}{\setminus} \cos \left(x\right)$ and $\setminus \tan \left(x\right) = \setminus \sin \frac{x}{\setminus} \cos \left(x\right)$, first multiply by $\setminus \cos \left(x\right)$. Thus

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

Put in #1=\sin^2(x)+\cos^2(x):

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

$\setminus {\cos}^{2} \left(x\right) + \setminus \sin \left(x\right) \setminus \cos \left(x\right) = 0$

$\setminus \cos \left(x\right) \left(\setminus \cos \left(x\right) + \setminus \sin \left(x\right)\right) = 0$

Thus the equation is true only if $\setminus \cos \left(x\right) = 0$ or $\setminus \cos \left(x\right) = - \sin \left(x\right)$.

But we cannot accept $\setminus \cos \left(x\right) = 0$ because that leaves $\setminus \sec \left(x\right) = \frac{1}{\setminus} \cos \left(x\right)$ undefined. So $\setminus \cos \left(x\right) = - \sin \left(x\right)$, thus $x = - \setminus \frac{\pi}{4}$ plus multiples of $\setminus \pi$.