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Home Q & A Board Science-Related Homework Help Mathematics
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lgorman20101 lgorman20101
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12 years ago
It is given that tan 3x = k tan x, where k is a constant and tan x is not = 0

By first expending tan(2x + x),

show that (3k -1)(tan^2 x) = k - 3
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#1
12 years ago
tan3x=tan(2x+x)=(tan2x+tanx)/(1+tan2xtanx)  and tan2x=(2tanx)/(1+tan^2x)

substitute these values  remembering that tan3x=ktanx

tan3x=ktanx=(2tanx/(1+tan^2x+tanx))/(1+2tan^2x)/(1+tan^2x)=(tanx(3+tan^2x)(1+3tan^2x)

divide both sides by tanx (possible since tanx is NOT = 0)

(3+tan^2x)/(1+3tan^2x)=k

cross multiply

k+3ktan^2x=3+tan^2x

3ktan^2x-tan^2x=3-k

tan^2x(3k-1)=3-k  (I get 3-k  and NOT k-3!)
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