# How do I show that cos theta/ (1 + sin theta) + (1+ sin theta)/cos theta = 2 sec theta ?

Sep 20, 2015

Yes, the equality is correct. See explanation.

#### Explanation:

$\frac{\cos \left(\theta\right)}{1 + \sin \left(\theta\right)} + \frac{1 + \sin \left(\theta\right)}{\cos \left(\theta\right)}$
Multiply the left fraction by $\cos \left(\theta\right)$ and the right fraction by $\left(1 + \sin \left(\theta\right)\right)$ to put everything under the same denominator:

$= \frac{{\left(\cos \left(\theta\right)\right)}^{2} + {\left(1 + \sin \left(\theta\right)\right)}^{2}}{\left(1 + \sin \left(\theta\right)\right) \cdot \cos \left(\theta\right)}$

Then we open the parentheses (at the numerator, just leave the denominator untouched) and get:
$= \frac{{\cos}^{2} \left(\theta\right) + 1 + 2 \sin \left(\theta\right) + {\sin}^{2} \left(\theta\right)}{\left(1 + \sin \left(\theta\right)\right) \cdot \cos \left(\theta\right)}$

You should remember the identity ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$ which you can see in the numerator. So we get:
$= \frac{1 + 2 \sin \left(\theta\right) + 1}{\left(1 + \sin \left(\theta\right)\right) \cdot \cos \left(\theta\right)}$

which simplifies to:
$= \frac{2 + 2 \sin \left(\theta\right)}{\left(1 + \sin \left(\theta\right)\right) \cdot \cos \left(\theta\right)}$

but we can factor out the $2$ so:
$= 2 \cdot \frac{1 + \sin \left(\theta\right)}{\left(1 + \sin \left(\theta\right)\right) \cdot \cos \left(\theta\right)}$

Now remove the $\left(1 + \sin \left(\theta\right)\right)$ from top (numerator) and bottom (denominator) (you can do this because any number over itself is 1).
Then you are left with:
$= 2 \cdot \frac{1}{\cos \left(\theta\right)}$
and since $\frac{1}{\cos \left(\theta\right)} = \sec \left(\theta\right)$ by definition of the secant you do indeed have:
$= 2 \cdot \sec \left(\theta\right)$

Q.E.D.