How do you solve $Tan^2(x)-|Sec(x)|=1$ ?

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Answer 1 · 2016-11-30T12:42:29

$x= 2kpi+-pi/3$ and

$x= 2kpi+-2/3pi, k = 0, +-1, +-2, +-3, ..$

$|sec x| >=0$ .

As $tan^2x = sec^2x-1 =|sec x|^2-1.$

the given equation becomes a quadratic in $|sec x|$ .

$|sec x|^2-|sec x|-2=(|sec x|-2)(|xec x|+1)=0$ .

So, $|sec x|=2 to sec x = +- 2$

$sec^(-1)(+-2)= pi/3 and 2/3pi$ .. And so,

$x= 2kpi+-pi/3$ and

$x= 2kpi+-2/3pi, k = 0, +-1, +-2, +-3, ..$

How do you solve Tan^2(x)-|Sec(x)|=1 Tan^2(x)-|Sec(x)|=1 ?