How do you use the sum and difference formula to simplify #(1-tan40tan20)/(tan40+tan20)#?

1 Answer
George C.
Sep 10, 2016

#(1 - tan 40 tan 20) / (tan 40 + tan 20) = cot 60#

Explanation:

The basic sum and difference formulae for #sin# and #cos# are:

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

#sin(alpha - beta) = sin alpha cos beta - sin beta cos alpha#

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

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

There are also sum and difference formulae for #tan# that we can derive from the above:

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

Similarly:

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

So notice that:

#(1 - tan 40 tan 20) / (tan 40 + tan 20) = 1/(tan (40+20)) = 1/tan 60 = cot 60#

Related questions
See all questions in Sum and Difference Identities
Impact of this question
3145 views around the world
You can reuse this answer
Creative Commons License