SOLUTION: prove the identity (cosθ - sinθ)/(cosθ + sinθ) = sec2θ - tan2θ

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Question 687496: prove the identity (cosθ - sinθ)/(cosθ + sinθ) = sec2θ - tan2θ
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%28cos%28theta%29+-+sin%28theta%29%29+%2F+%28cos%28theta%29+%2B+sin%28theta%29%29%22%22=%22%22sec%282theta%29+-+tan%282theta%29

          = 1%2F%28cos%282theta%29%29%22%22-%22%22sin%282theta%29%2Fcos%282theta%29

          = %281-sin%282theta%29%29%2Fcos%282theta%29

Use the identities  sin%282alpha%29+=+2sin%28alpha%29cos%28alpha%29
                    cos%282alpha%29+=+cos%5E2%28alpha%29-sin%5E2%28alpha%29

          = %281-2sin%28theta%29cos%28theta%29%29%2F%28cos%5E2%28theta%29-sin%5E2%28theta%29%29

The trick here is to replace the 1 by cosē(theta)+sinē(theta),
so the numerator will become factorable after
rearranging the terms, and we see that the denominator
is factorable as the difference of squares:

          = 

Rearrange the terms and the order of factors in the numerator
to show that it is factorable:

          = 

Factor numerator and denominator:

          = 

          = 

          = %28cos%28theta%29+-+sin%28theta%29%29+%2F+%28cos%28theta%29+%2B+sin%28theta%29%29

Edwin