# `cos(x) - cos(x)/(1 - tan(x)) = (sin(x)cos(x))/(sin(x) - cos(x))` Verify the identity.

mathace | (Level 3) Assistant Educator

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Verify the identity. `cos(x)-[cos(x)/(1-tan(x))]=(sinxcosx)/(sinx-cosx)`

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

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

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

Rewrite the  tan(x) terms as the quotient sin(x)/cos(x).

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

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

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

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

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

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

balajia | College Teacher | (Level 1) eNoter

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By cross multiplication,we get

`cosx-(cosx/(1-tanx)) = (cosx(1-tanx)-cosx)/(1-tanx)`

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

`=(-sinx)/(1-tanx)`

`=(-sinx.cosx)/(cosx-sinx)`

`=(sinx.cosx)/(sinx-cosx)`

scisser | (Level 3) Honors

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`cosx - [(cosx)/(1 -tanx)] = (sinxcosx)/(sinx - cosx) `
LCD is (1 - tanx)..

`( [cosx(1 - tanx)] - cosx ) / (1 - tanx) = (sinxcosx)/(sinx - cosx) `

`(cosx - cosxtanx - cosx)/ (1 - tanx) = (sinxcosx)/(sinx - cosx) `

`(-cosxtanx) / (1 - tanx) = (sinxcosx)/(sinx - cosx) `

but `(-cosxtanx) = - cosx (sinx / cosx) = sinx `

so our equation becomes...

`-sinx / (1 - tanx) = (sinxcosx) / (sinx - cosx) `
` -sinx / [1 - (sinx / cosx) ] = (sinxcosx) / (sinx - cosx) `

`-sinx / [ (cosx - sinx) / cosx ] = (sinxcosx) / (sinx - cosx) `

`-(sinxcosx) / (cosx - sinx) = (sinxcosx) / (sinx - cosx) `

`(sinxcosx) / (sinx - cosx) = (sinxcosx) / (sinx - cosx) `