# How do you verify the identity (1+sintheta)/costheta+costheta/(1+sintheta)=2sectheta?

Feb 6, 2017

see below

#### Explanation:

Left Hand Side:

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

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

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

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

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

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

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

$= \frac{2}{\cos} \theta$

$= 2 \cdot \frac{1}{\cos} \theta$

$= 2 \sec \theta$

$\therefore =$ Right Hand Side