How do you solve $sec(x)-1=tan(x)$ ?

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How do you solve $sec(x)-1=tan(x)$ ?

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Answer 1 · 2018-03-06T14:14:55

$x=2kpi,kinZ$

Explanation:

$secx-1=tanxrArr1/cosx-1=sinx/cosxrArr1-cosx=sinxrArrcosx+sinx=1rArr1/sqrt(2)cosx+1/sqrt(2)sinx=1/sqrt(2)$
$rArrcosxcos(pi/4)+sinxsin(pi/4)=cos(pi/4)$
$rArrcos(x-pi/4)=cos(pi/4)$
$x-pi/4=2kpi+-pi/4,kinZ$
$x=2kpi+-pi/4+pi/4,kinZ$
$x=2kpi+pi/4+pi/4,kinZorx=2kpi-pi/4+pi/4.kinZ$
$color(red)(x=2kpi+pi/2,kinZorx=2kpi,kinZ)$
But, taking $k=0,..etc.$
$x=2kpi+pi/2.kinZ=>x=pi/2$ , which does not satisfy
$secx-1=tanx$ , as $sec (pi/2) and tan(pi/2)$ are undefined.
So, $color(red)(x!=2kpi+pi/2,kinZrArrx=2kpi,kinZ)$

Answer 2 · 2018-03-06T14:22:36

$x = 2kpi , "where k any integer"$

Explanation:

$sec(x)−1=tan(x)$

$1/cosx −1=sinx/cosx$

$(1-sinx)/cosx =1$

$sinx + cosx = 1, "where" cosx ≠ 0$

$x = 2kpi , "where k any integer"$

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How do you solve $sec(x)-1=tan(x)$ ?

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