SOLUTION: How do I prove that {{{(sinC+cosC-1)/(sinC-cosC+1)=cosC/(sinC+1)}}}?

Algebra ->  Trigonometry-basics -> SOLUTION: How do I prove that {{{(sinC+cosC-1)/(sinC-cosC+1)=cosC/(sinC+1)}}}?      Log On


   

Question 1106478: How do I prove that %28sinC%2BcosC-1%29%2F%28sinC-cosC%2B1%29=cosC%2F%28sinC%2B1%29?
Found 2 solutions by Edwin McCravy, Alan3354:
Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!


%28sin%28C%29-1%2Bcos%28C%29%29%2F%28sin%28C%29%2B1-cos%28C%29%29

%28%28sin%28C%29-1%29%2Bcos%28C%29%29%2F%28%28sin%28C%29%2B1%29-cos%28C%29%29%22%22%2A%22%22%28%28sin%28C%29%2B1%29%2Bcos%28C%29%29%2F%28%28sin%28C%29%2B1%29%2Bcos%28C%29%29

Let's multiply the numerators first, then we'll multiply
the denominators, then we'll put them together as a fraction:



use "FOIL":





Simplify

sin%5E2%28C%29%2Bcos%5E2%28C%29-1%2B2cos%28C%29sin%28C%29

1-1%2B2cos%28C%29sin%28C%29

2cos%28C%29sin%28C%29  <-- simplified numerator

-----

Now let's multiply the denominators:



use "FOIL":



sin%5E2%28C%29%2B2sin%28C%29%2B1-cos%5E2%28C%29

sin%5E2%28C%29%2B2sin%28C%29%2B1-%281-sin%5E2%28C%29%29

sin%5E2%28C%29%2B2sin%28C%29%2B1-1%2Bsin%5E2%28C%29

2sin%5E2%28C%29%2B2sin%28C%29  

2sin%28C%29%28sin%28C%29%5E%22%22%2B1%29  <-- simplified denominator

Finally put simplified numerator over simplified denominator:

%282cos%28C%29sin%28C%29%29%2F%282sin%28C%29%28sin%28C%29%5E%22%22%2B1%29%29



cos%28C%29%2F%28sin%28C%29%5E%22%22%2B1%29%29

Edwin

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How do I prove that %28sinC%2BcosC-1%29%2F%28sinC-cosC%2B1%29=cosC%2F%28sinC%2B1%29?
-----
Cross multiply.
sin^2 + sin*cos - sin + sin + cos - 1 = sin*cos - cos^2 + cos
sin^2 - 1 = - cos^2
sin^2 + cos^2 = 1
I'm convinced.
====================
Someone might comment on not working on just 1 side.
I've never seen an example where that makes a difference.