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If tan 2x =2 for 0 < x < 90 , determine the exact value of tan x !
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\[Tan((Tan^{-1}2)\div2)\]
tan 2x = \(\dfrac{2tanx}{1-tan^2x}=2\) \(\rightarrow 2tanx = 2(1-tan^2x)=2-2tan^2x\) so, tan^2 +tan x -1 =0 solve as a quadratic by let tan x =t, you have t^2 +t -1 =0 after solving for t, plug back to get tan x
ok, thanks
np
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