# Prove cosx/(1-tanx) - cosx = sin x - sinx/(1-cotx)

hala718 | High School Teacher | (Level 1) Educator Emeritus

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cosx/(1-tanx) - cosx = sinx-sinx/(1-cotx)

Let us group similar terms:

==> cosx/(1-tanx) + sinx/(a-cotx) = sinx + cosx

We know that :

tanx = sinx/cosx

cotx = cosx/sinx

==> cosx/(1- sinx/cosx) + sinx/(1- cosx/sinx) = sinx + cosx

Now let us simplify:

==> cosx/[(cosx-sinx)/cosx]+ sinx/[(sinx-cosx)/sinx]= sinx+cosx

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

Let us rewrite" sin^2x/(sinx-cosx) = -sin^2 x/(cosx-sinx)

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

We know that:

a^2 - b^2 = (a-b)(a+b)

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

Reduce similar terms:

==> cosx + sinx = sinx + cosx

Then the equality is true.

neela | High School Teacher | (Level 3) Valedictorian

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To prove cosx/(1-tanx) -cosx = sinx - sinx(1-cotx)

We simplify LHS:

cosx/(1-tanx) -cosx = cosx/(1-sinx/cosx)  - cosx.

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

= sinxcosx/(cosx-sinx) , as other terms in numerator gets cancelled.

= sinxcosx/(cosx-sinx).........(1)

RHS :

sinx - sinx/(1-cotx) = sinx - sinx/(1- cosx/sinx)

= sinx - sin^2x(sinx-cosx)

= {sinx(sinx-cosx) -sin^2x}/((sinx-cosx)

= -cosxsinx/(sinx-cosx)

= sinxcosx/(cosx-sinx).........(2)

From (1) and (2) we conclide that LHS and RHS are same. Therefore the given identity  cosx/(1-tanx) - cosx = sinx - sinx/(1-cotx) is established.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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