Question

How do you simplify `(tan 4Θ + tan 2Θ) / (1-tan 4Θtan 2Θ)`?

Answer 1

`(tan 4 theta + tan 2 theta)/(1 - tan 4 theta tan 2 theta) = tan 6 theta`

Explanation:

The sum of angles formulae for `sin` and `cos` can be written:

`sin (alpha+beta) = sin alpha cos beta + sin beta cos alpha`

`cos (alpha + beta) = cos alpha cos beta - sin alpha sin beta`

Hence we find:

`tan (alpha + beta) = (sin (alpha + beta))/(cos (alpha + beta))`

`color(white)(tan (alpha + beta)) = (sin alpha cos beta + sin beta cos alpha)/(cos alpha cos beta - sin alpha sin beta)`

`color(white)(tan (alpha + beta)) = ((sin alpha cos beta + sin beta cos alpha) -: (cos alpha cos beta))/((cos alpha cos beta - sin alpha sin beta) -: (cos alpha cos beta))`

`color(white)(tan (alpha + beta)) = (sin alpha / cos alpha + sin beta / cos beta)/(1 - sin alpha / cos alpha sin beta / cos beta)`

`color(white)(tan (alpha + beta)) = (tan alpha + tan beta)/(1 - tan alpha tan beta)`

So the sum of angles formula for `tan` may be written:

`tan (alpha + beta) = (tan alpha + tan beta)/(1- tan alpha tan beta)`

Putting `alpha=4 theta` and `beta = 2 theta`, we find:

`(tan 4 theta + tan 2 theta)/(1 - tan 4 theta tan 2 theta) = tan (4 theta + 2 theta) = tan 6 theta`