Question

Help w/ a trigonometric problem?

Answer 1

Sure!

Answer 2

\[\frac{ 1 - 3 \sin \theta }{ 1 - \sin \theta } = \frac{ 1 - 2\sin \theta - 3\sin ^2\theta }{ \cos^2\theta } \]

Thanks!

Answer 3

Give me a second.

Answer 4

Uh, are you supposed to solve for something or just see if the results are equal?

Answer 5

You're supposed to verify, so choose a side and get it equal to the other,

Answer 6

I'm not really sure how to do this. @Agl202

Answer 7

I do not believe these 2 sides are equal, I've tried a few things and can not find a way to verify

Answer 8

\[ \frac{1-2\sin(\theta)-3\sin^2(\theta)}{\cos^2(\theta)} \]
\[= \frac{1-3\sin(\theta) + \sin(\theta) - 3\sin^2(\theta)}{1-\sin^2(\theta)}\]
\[= \frac{[1-3\sin(\theta)]\cdot[1+\sin(\theta)]}{[1-\sin(\theta)]\cdot[1+\sin(\theta)]}\]
\[= \frac{1-3\sin(\theta)}{1-\sin(\theta)}\]

Answer 9

Thank you so much, I actually understand that!