#tan(60+x)/tan(60-x)color(white)("d")# use the formula #cosA cosB-sinA sinB# ?

1 Answer
Sonnhard
Jun 1, 2018

#(sqrt(3)+4tan(x)+sqrt(3)tan^2(x))/(sqrt(3)-4tan(x)-sqrt(3)tan^2(x))#

Explanation:

Using that
#tan(pi/3+x)=(tan(pi/3)+tan(x))/(1-tan(pi/3)tan(x))#
and

#tan(pi/3-x)=(tan(pi/3)-tan(x))/(1+tan(pi/3)*tan(x))#
and

#tan(pi/3)=sqrt(3)#
so we have
#((sqrt(3)+tan(x))*(1+sqrt(3)tan(x)))/((1-sqrt(3)tan(x))(sqrt(3)-tan(x)))#
multiplying out we get

#(sqrt(3)+4tan(x)+sqrt(3)tan^2(x))/(sqrt(3)-4tan(x)-sqrt(3)tan^2(x))#

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