How do you simplify the expression (sintheta+tantheta)/(1+sectheta)?

Sep 9, 2016

$\sin \left(\theta\right)$

Explanation:

We have: $\frac{\sin \left(\theta\right) + \tan \left(\theta\right)}{1 + \sec \left(\theta\right)}$

Let's apply two standard trigonometric identities; $\tan \left(\theta\right) = \frac{\sin \left(\theta\right)}{\cos \left(\theta\right)}$ and $\sec \left(\theta\right) = \frac{1}{\cos \left(\theta\right)}$:

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

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

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

Let's apply the Pythagorean identity ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$:

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

We can rearrange the Pythagorean identity to get:

$\implies {\sin}^{2} \left(\theta\right) = 1 - {\cos}^{2} \left(\theta\right)$

Let's apply this to get:

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

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

$= \sin \left(\theta\right)$