# trigonometryDemonstrate if cosx/(1-tanx) - cosx=sin x - sinx/(1-cotx).

sciencesolve | Certified Educator

You need to bring the terms to the left to a common denominator such that:

(cosx - cos x + cos x*tan x)/(1-tanx) =sin x - sinx/(1-cotx)

Reducing like terms yields:

(cos x*tan x)/(1-tanx) =sin x - sinx/(1-cotx)

You should remember that tan x = sin x/cos x, hence cos x*tan x = sin x.

sin x/(1-tanx) = sin x - sinx/(1-cotx)

You should  bring the terms to the right to a common denominator such that:

sin x/(1-tanx) = (sin x - sin x*cot x - sin x)/(1-cot x)

sin x/(1-tanx) = -cos x/(1-cot x)

-cos x(1-tan x) = sin x(1 - cot x)

-cos x + cos x*tan x = sin x - sin x*cot x

-cos x + cos x*sin x/cos x = sin x - sin x*cos x/sin x

-cos x + sin x = sin x - cos x

Hence, the last line proves that cosx/(1-tanx) - cosx=sin x - sinx/(1-cotx).

justaguide | Certified Educator

We have to show that cos x/(1-tan x) - cos x=sin x - sin x/(1-cot x)

cos x/(1-tan x) - cos x

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

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

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

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

=> [(cos x * sin x]/(cos x - sin x) ...(1)

sin x - sin x/(1-cot x)

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

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

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

=> (sin x)(cos x)/(cos x - sin x) ...(2)

As (1) = (2) we have cos x/(1-tan x) - cos x=sin x - sin x/(1-cot x)

giorgiana1976 | Student

For the beginning, we'll re-arrange the terms of the given expression:

cos x/(1-tan x) - cos x = sin x - sin x/(1-cot x)

sin x + cos x = cos x/(1-tan x) + sin x/(1-cot x)

We'll re-write the terms from the left side of the given expression:

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

cos x/(1- sin x/cos x) = cos x/[(cos x-sin x)/cos x]

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

We'll re-write the terms from the right side of the given expression:

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

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

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

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

We'll re-write the difference of squares:

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

The left side of the expression will become:

(cos x - sin x)(cos x + sin x)/(cos x-sin x)

We'll reduce the like terms:

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