005 Sample Final A, Question 14

Question Prove the following identity,

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

ExpandFoundations:
1) What can you multiply $1-\sin(\theta )$ by to obtain a formula that is equivalent to something involving $\cos$?
Answers:
1) You can multiply $1-\sin(\theta )$ by Failed to parse (syntax error): {\displaystyle \frac{1 + \sin(\theta)}{\1 + \sin(\theta)}}

ExpandStep 1:
We start with the left hand side. We have ${\frac {1-\sin(\theta )}{\cos(\theta )}}={\frac {1-\sin(\theta )}{\cos(\theta )}}{\Bigg (}{\frac {1+\sin(\theta )}{1+\sin(\theta )}}{\Bigg )}$.

ExpandStep 2:
Simplifying, we get ${\frac {1-\sin(\theta )}{\cos(\theta )}}={\frac {1-\sin ^{2}(\theta )}{\cos(\theta )(1+\sin(\theta ))}}$.

ExpandStep 3:
Since $1-\sin ^{2}(\theta )=\cos ^{2}(\theta )$, we have
${\frac {1-\sin(\theta )}{\cos(\theta )}}={\frac {\cos ^{2}(\theta )}{\cos(\theta )(1+\sin(\theta ))}}={\frac {\cos(\theta )}{1+\sin(\theta )}}$