Quote History Quoted:
I think that I have number 1 can someone please check.
step 1) cosx/1+sinx (sin x is the same as 1+sin x so I multiply times the recipical 1-sin x)
step 2) cosx (1-sinx)/(1+sinx)(1-sinx) which breaks down into cosx(1-sinx)/(1-sinx^2)
step 3) because 1-sinx^2 is the same as 1+cos^2 I have 1+cos^2 which foiled out is (1+cos)(1+cos) so now my problem looks like thus cosx(1-sinx)/(1+cosx)(1+cosx)
step 4) the cosx on the numerator and denominator cancel out and I am left with (1-sinx)/(1+cosx) which is the same as (1/cosx) - (sinx/cosx) which equals (secx)-(tanx)
correct?
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No.
Your math has a series of errors. (1 + cos X)^2 does not = 1 + cos^2 x, but rather = 1 + 2 (cos x) + cos^2 x. Also, there is no cos x in the denominator to cancel out.
Solution:
Starting with the initial equation:
cosx/(1+sinx) = secx-tanx
rewrite everything in terms of cos and sin:
cos x / (1 + sin x) = (1 /cos x) - (sin x / cos x)
simplifying the right side becomes (1 - sin x)/ cos x
so:
cos x / (1 + sin x) = (1 - sin x) / cos x
multiply both sides by (cos x) (1 + sin x) to get the same denominators, and simplifying
gives us
cos x * cos x = (1 + sin x) (1 - sin x)
cos^2 x = 1 - sin^2 x
and then adding sin^2x to both sides
cos^2 x + sin^2 x = 1
since this is true, we prove the original equality.