# cosx/1-sinx=1+cosx+sinx/1+cosx-sinx?

### 5 Answers

- PhilipLv 62 weeks ago
I interpret your equation to be cosx/(1-sinx) = (1+cosx+sinx)/(1+cosx-sinx). I do wish

someone would teach you to use brackets properly so one doesn't have to guess at

what you mean!.

Introduce sane notation as follows: Put (s,c) = (sinx,cosx).

Then your equation becomes c/(1-s) = (1+c+s)/(1+c-s)...(1). { neat or what?}.

Multiply (1) through by (1-s)(1+c-s).

Then c(1+c-s) = (1-s)(1+c+s), ie., c + c^2 -sc = 1+c+s -s-sc-s^2.

Add sc-c to both sides.

Then c^2 = 1-s^2, ie., c^2+s^2 = 1, which is an identity valid for all real x

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- MyRankLv 62 months ago
cosx/1-sinx = 1+cosx+sinx / 1+cosx – sinx

L.H.S cosx/1-sinx

cos²x/2 - sin²x/2 / sin²x/2 + cosx/2 -2sinx/2cosx/2

= (cosx/2 – sinx/2) (cosx/2 + sinx/2) / (sinx/2 – cosx/2)²

= (cosx/2 + sinx/2) / (sinx/2 - cosx/2)

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- KrishnamurthyLv 72 months ago
cosx/(1 - sinx) =

(1 + sinx)cosx/(1 - sin^2x)

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- PopeLv 72 months ago
Order of operations. Learn them.

If I tried to guess what that actually meant, I might come up with ten different interpretations. Let me take you at your word and interpret it literally.

cosx/1-sinx=1+cosx+sinx/1+cosx-sinx

cos(x) - sin(x) = 1 + cos(x) + sin(x) + cos(x) - sin(x)

cos(x) - sin(x) = 1 + 2cos(x)

sin(x) + cos(x) = -1

1/√(2)sin(x) + 1/√(2)cos(x) = -1/√(2)

sin(x)cos(π/4) + cos(x)sin(π/4) = -1/√(2)

sin(x + π/4) = -1/√(2)

sin(x + π/4) = sin(-π/4)

x + π/4 = kπ + (-1)ᵏ(-π/4)

x = kπ - π/4 + (-1)ᵏ(-π/4)

x = [4k - 1 - (-1)ᵏ]π/4, where k is any integer

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- 2 months ago
(1 + cos(x) + sin(x)) / (1 + cos(x) - sin(x)) =>

(1 + cos(x) + sin(x)) * (1 + cos(x) + sin(x)) / ((1 + cos(x) - sin(x)) * (1 + cos(x) + sin(x))) =>

((1 + cos(x))^2 + 2 * sin(x) * (1 + cos(x)) + sin(x)^2) / ((1 + cos(x))^2 - sin(x)^2) =>

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

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

(2 + 2sin(x) + 2cos(x) + 2sin(x)cos(x)) / (2cos(x) + 2cos(x)^2) =>

2 * (1 + sin(x)) * (1 + cos(x)) / (2 * cos(x)) * (1 + cos(x))) =

(1 + sin(x)) / cos(x) =>

(1 - sin(x)^2) / (cos(x) * (1 - sin(x))) =>

cos(x)^2 / (cos(x) * (1 - sin(x))) =>

cos(x) / (1 - sin(x))

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