# 005 Sample Final A, Question 14

Question Prove the following identity,

${\displaystyle {\frac {1-\sin(\theta )}{\cos(\theta )}}={\frac {\cos(\theta )}{1+\sin(\theta )}}}$

Foundations:
1) What can you multiply ${\displaystyle 1-\sin(\theta )}$ by to obtain a formula that is equivalent to something involving ${\displaystyle \cos }$?
2) You can multiply ${\displaystyle 1-\sin(\theta )}$ by ${\displaystyle {\frac {1+\sin(\theta )}{1+\sin(\theta )}}}$
We start with the left hand side. We have ${\displaystyle {\frac {1-\sin(\theta )}{\cos(\theta )}}={\frac {1-\sin(\theta )}{\cos(\theta )}}{\Bigg (}{\frac {1+\sin(\theta )}{1+\sin(\theta )}}{\Bigg )}}$.
Simplifying, we get ${\displaystyle {\frac {1-\sin(\theta )}{\cos(\theta )}}={\frac {1-\sin ^{2}(\theta )}{\cos(\theta )(1+\sin(\theta ))}}}$.
Since ${\displaystyle 1-\sin ^{2}(\theta )=\cos ^{2}(\theta )}$, we have
${\displaystyle {\frac {1-\sin(\theta )}{\cos(\theta )}}={\frac {\cos ^{2}(\theta )}{\cos(\theta )(1+\sin(\theta ))}}={\frac {\cos(\theta )}{1+\sin(\theta )}}}$