Question 105525
Let {{{y=(tan x)^2}}}


Then,

{{{y^2+2y+1=0}}}


Please read my lesson,"Factoring trinomials: SHORTCUT!!!"


By the quadratic formula,

{{{y=(-2+-sqrt(2^2-4*1*1))/(2*1)}}}
{{{y=(-2+-sqrt(4-4))/(2)}}}
{{{y=(-2+-sqrt(0))/(2)}}}
{{{y=(-2+- 0)/(2)}}}
{{{y=-2/2}}}
{{{y=-1}}}

Thus,
{{{(tan x)^2=-1}}}
{{{(tan x)^2+1=0}}}

So, this means that the equation is a perfect square trinomial and that:


{{{(tan x)^4+2(tan x)^2+1=((tan x)^2+1)^2}}}


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HyperBrain!