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Determine the value of a: 2tanx - tan2x + 2a = 1 - (tan2x)(tan^2x)
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I have no idea if i did it right but i got: a = (1-tanx)/2
The answer is 1/2
I have no idea how to get the answer tho :(
\[2\tan x-\tan 2x+2a=1-\left( \tan 2x \right)\left( \tan ^2x \right)\] \[\tan 2x \tan ^2x-\tan 2x+2a+2\tan x-1=0\] \[\tan 2x \left( \tan ^2x-1 \right)+2a=1- 2\tan x\] \[\frac{ 2\tan x }{ 1-\tan ^2x }\left\{ -\left( 1-\tan ^2x \right) \right\}+2a=1-2\tan x\] \[-2\tan x+2a=1-2\tan x\] a=1/2, assume \[1-\tan ^2x \neq 0\]
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