SOLUTION: How do I verify the trig identity? (1+sinx)^2/cos^2(x) = (1 + sinx)/ (1-sinx )

Question 743475: How do I verify the trig identity?
(1+sinx)^2/cos^2(x) = (1 + sinx)/ (1-sinx )
Found 3 solutions by jim_thompson5910, Alan3354, tommyt3rd:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Pick one side and transform it into the other. Once you've chosen a side, you cannot alter the other side at all. I'm going to pick the right side and transform it into the left side.

To do that, multiply top and bottom of the fraction by 1+sin(x) and simplify

[ (1 + sin(x))^2 ]/(cos^2(x)) = (1+sin(x))/(1-sin(x))

[ (1 + sin(x))^2 ]/(cos^2(x)) = [(1+sin(x))(1+sin(x))]/[(1-sin(x))(1+sin(x))]

[ (1 + sin(x))^2 ]/(cos^2(x)) = [(1+sin(x))^2]/[1^2-sin^2(x)]

[ (1 + sin(x))^2 ]/(cos^2(x)) = [(1+sin(x))^2]/[1-sin^2(x)]

[ (1 + sin(x))^2 ]/(cos^2(x)) = [(1+sin(x))^2]/(cos^2(x))

Both sides are now identical, so this verifies the identity.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
verify the trig identity?
(1+sinx)^2/cosx^2 = 1 + sinx/ 1-sinx
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(1+sinx)^2/(1 - sin^2) = 1 + sinx/ 1-sinx
(1+sinx)^2/((1 - sin)*(1 + sin)) = 1 + sinx/ 1-sinx
(1+sinx)^2/(1 + sin)) = 1 + sinx
1 + sin = 1 + sin
---------------------
There's no procedure, no answer to "How do I... ?"
You try things.
Having someone do it for you doesn't accomplish anything.
Answer by tommyt3rd(5050)   (Show Source): You can put this solution on YOUR website!

For convenience we'll denote sin(x) as sinx. First we expand the numerator and rewrite the denominator:

Using the difference of squares identity for the denominator...

after cancelling...

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