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

1 Answer
George C.
Oct 9, 2016

#(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#

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