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

Feb 8, 2016

I strongly assume that you would like to prove the identity

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

without the $x$ between $\sin \left(x\right)$ and $\tan \left(x\right)$.
(Maybe it was a "times" that was confused with "$x$" on a hand-written note).

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We will need the following identities:

 $\text{ } \tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

 $\text{ } \sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

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

Let's start at the left side and try to get to the right side:

$\sin \left(x\right) \tan \left(x\right) + \cos \left(x\right) \stackrel{\text{ }}{=} \sin \left(x\right) \cdot \sin \frac{x}{\cos} \left(x\right) + \cos \left(x\right)$

$= {\sin}^{2} \frac{x}{\cos} \left(x\right) + \cos \left(x\right) \cdot \textcolor{b l u e}{\cos \frac{x}{\cos} \left(x\right)}$

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

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

$\stackrel{\text{ }}{=} \frac{1}{\cos} \left(x\right)$

$\stackrel{\text{ }}{=} \sec \left(x\right)$

q.e.d.