Solve for $a$ : $2 tan x - tan 2x +2a=1-tan2x tan^2 x$ ?

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Answer 1 · 2017-11-22T17:05:04

$a=1/2.$

We know that, $tan2x=(2tanx)/(1-tan^2x)=(2t)/(1-t^2), where, t=tanx.$

$:. 2tanx-tan2x+2a=1-tan2xtan^2x,$

$rArr 2t-(2t)/(1-t^2)+2a=1-(2t)/(1-t^2)*t^2.$

Multiplying by $(1-t^2),$ we get,

$2t(1-t^2)-2t+2a(1-t^2)=(1-t^2)-2t^3.$

$:. cancel(2t-2t^3-2t)+2a(1-t^2)=1-t^2cancel(-2t^3), or,$

$2a(1-t^2)=(1-t^2).$

$:. 2a=(1-t^2)/(1-t^2)=1, if t^2ne1.$

$:. a=1/2, if t^2ne1.$

Now, $t^2=1 rArr t=tanx=+-1=tan(pmpi/4),$

$rArr x=npi+-pi/4, n in ZZ,$ but, then, $tan2x$ becomes undefined.

$:. t=tanxnepm1.$

$:. a=1/2.$

Solve for aa: 2 tan x - tan 2x +2a=1-tan2x tan^2 x2 tan x - tan 2x +2a=1-tan2x tan^2 x ?