005 Sample Final A, Question 14

Question Prove the following identity,

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

Foundations:
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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1 + \sin(\theta)}{\1 + \sin(\theta)}}

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

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

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

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