Question #19476

Feb 14, 2017

Explanation:

$\sec y + \tan y = \frac{1}{\cos} y + \sin \frac{y}{\cos} y = \frac{\cos y + \cos y \sin y}{\cos} ^ 2 y =$

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

Let's break this down:

1. Rewrite $\sec y$ and $\tan y$ as $\frac{1}{\cos} y$ and $\sin \frac{y}{\cos} y$
3. Take a factor of $\cos y$ from the numerator
4. Rewrite ${\cos}^{2} y$ as $1 - {\sin}^{2} y$ using ${\cos}^{2} y + {\sin}^{2} y = 1$
5. Rewrite $1 - {\sin}^{2} y$ using the difference of two squares identity