How do you simplify tan(x+y)tan(x+y) to trigonometric functions of x and y?

1 Answer
Jan 6, 2016

tan(x+y)=(tanx+tany)/(1-tanxtany)tan(x+y)=tanx+tany1tanxtany

Explanation:

This can be expanded through the tangent angle addition formula:

tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)tan(α+β)=tanα+tanβ1tanαtanβ

Thus,

tan(x+y)=(tanx+tany)/(1-tanxtany)tan(x+y)=tanx+tany1tanxtany


The tangent addition formula can be found using the sine and cosine angle addition formulas.

sin(alpha+beta)=sinalphacosbeta+cosalphasinbetasin(α+β)=sinαcosβ+cosαsinβ
cos(alpha+beta)=cosalphacosbeta-sinalphasinbetacos(α+β)=cosαcosβsinαsinβ

Since tanx=sinx/cosxtanx=sinxcosx,

tan(alpha+beta)=sin(alpha+beta)/cos(alpha+beta)=(sinalphacosbeta+cosalphasinbeta)/(cosalphacosbeta-sinalphasinbeta)tan(α+β)=sin(α+β)cos(α+β)=sinαcosβ+cosαsinβcosαcosβsinαsinβ

This can be written in terms of tangent by dividing both the numerator and denominator by cosalphacosbetacosαcosβ.

tan(alpha+beta)=((sinalphacosbeta+cosalphasinbeta)/(cosalphacosbeta))/((cosalphacosbeta-sinalphasinbeta)/(cosalphacosbeta))=(sinalpha/cosalpha(cosbeta/cosbeta)+sinbeta/cosbeta(cosalpha/cosalpha))/(cosalpha/cosalpha(cosbeta/cosbeta)-sinalpha/cosalpha(sinbeta/cosbeta))tan(α+β)=sinαcosβ+cosαsinβcosαcosβcosαcosβsinαsinβcosαcosβ=sinαcosα(cosβcosβ)+sinβcosβ(cosαcosα)cosαcosα(cosβcosβ)sinαcosα(sinβcosβ)

Final round of simplification yields:

tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)tan(α+β)=tanα+tanβ1tanαtanβ