# trying to figure out how ( cot x - tan x ) / ( cot x + 1 ) = 1 - tan x

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### 1 Answer

You need to remember that `tan x = sin x/ cos x` and `cot x = cos x/ sin x` such that:

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

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

You need to diagonal(cross) multiply such that:

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

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

Converting the product `((cos x + sin x)*(cos x - sin x))` into a difference of squares yields:

`((cos x + sin x)*(cos x - sin x)) = cos^2 x - sin^2 x`

`(cos^2 x - sin^2 x)/(sin x*cos x) = (cos^2 x - sin^2 x)/(sin x*cos x) `

**Hence, making the more helpful trigonometric substitutions yields that `(cos^2 x - sin^2 x)/(sin x*cos x) = (cos^2 x - sin^2 x)/(sin x*cos x), ` thus the identity `(cot x - tan x)/(cot x + 1) = 1 - tan x` holds.**