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

Question

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

1 Answer

Impact of this question

67282 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-06T02:57:38

$tan (x + y) = \frac{tan x + tan y}{1 - tan x tan y}$ $tan(x+y)=(tanx+tany)/(1-tanxtany)$

Explanation:

This can be expanded through the tangent angle addition formula:

$tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β}$ $tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)$

Thus,

$tan (x + y) = \frac{tan x + tan y}{1 - tan x tan y}$ $tan(x+y)=(tanx+tany)/(1-tanxtany)$

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

$sin (α + β) = sin α cos β + cos α sin β$ $sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta$
$cos (α + β) = cos α cos β - sin α sin β$ $cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta$

Since $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ ,

$tan (α + β) = \frac{sin (α + β)}{cos (α + β)} = \frac{sin α cos β + cos α sin β}{cos α cos β - sin α sin β}$ $tan(alpha+beta)=sin(alpha+beta)/cos(alpha+beta)=(sinalphacosbeta+cosalphasinbeta)/(cosalphacosbeta-sinalphasinbeta)$

This can be written in terms of tangent by dividing both the numerator and denominator by $cos α cos β$ $cosalphacosbeta$ .

$tan (α + β) = \frac{\frac{sin α cos β + cos α sin β}{cos α cos β}}{\frac{cos α cos β - sin α sin β}{cos α cos β}} = \frac{\frac{sin α}{cos α} (\frac{cos β}{cos β}) + \frac{sin β}{cos β} (\frac{cos α}{cos α})}{\frac{cos α}{cos α} (\frac{cos β}{cos β}) - \frac{sin α}{cos α} (\frac{sin β}{cos β})}$ $tan(alpha+beta)=((sinalphacosbeta+cosalphasinbeta)/(cosalphacosbeta))/((cosalphacosbeta-sinalphasinbeta)/(cosalphacosbeta))=(sinalpha/cosalpha(cosbeta/cosbeta)+sinbeta/cosbeta(cosalpha/cosalpha))/(cosalpha/cosalpha(cosbeta/cosbeta)-sinalpha/cosalpha(sinbeta/cosbeta))$

Final round of simplification yields:

$tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β}$ $tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

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