Quote History Originally Posted By chenault:

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?

View Quote

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.