# Prove that (cosx/1-tanx) + (sinx/1-sinx) = sinx + cosx

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The trigonometric identity `cos x/(1-tanx) + sin x/(1-sinx) = sin x + cos x` has to be proved.

If x = 45 degrees, the left hand side is equal to `oo` as `(1/sqrt 2)/(1 - 1) = 1/0 = oo` , but the right hand side is equal to `sqrt 2`

The given equation is not a trigonometric identity.

If the `1 - sin x` in the term `sin x/(1 - sin x)` is replaced with `1 - cot x` we have:

`cos x/(1-tanx) + sin x/(1-cot x)`

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

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

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

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

= `cos x + sin x`

**The equation `cos x/(1-tanx) + sin x/(1-sinx) = sin x + cos x` is not a trigonometric identity. `cos x/(1-tanx) + sin x/(1-cot x) = sin x + cos x` on the other hand is a trigonometric identity.**

Thanks for pointing out the error. That had me stumped.

Test, ERRATA CORRIGE: `x=kpi, x=128^o10'21''`

for `x=kpi, sinx=0, cosx=+-1, tanx=0`

thwen we have: `+-1=+-1`

in the second root case is harder to.``

On the other side, if it was an equation could find some value wich hold true:

``

``

Thus we have:

`[sinx(1-tanx)+(1-sinx)]/[(1-sinx)(1-tanx)]=0`

that becames:

`[1-(sin^2x)/(cosx)]/[(1-sinx)(1-tanx)]=0`

Now, to find solution is enough find x s.t:

`sin^2x=cosx`

Developing:

`cos^2x+cosx-1=0`

`(cosx+1/2)^2-5/4=0`

`cos x+1/2= +-1/2sqrt(5)`

`` `cosx=1/2(-1+- sqrt(5))`

`x=128^o10'21''`

Then solution are : `x=0 , x=128^o10'21''`

The test on the next....