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

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

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Answer 1 · 2018-03-06T14:14:55

$x = 2 k π, k \in Z$ $x=2kpi,kinZ$

Explanation:

$sec x - 1 = tan x \Rightarrow \frac{1}{cos x} - 1 = \frac{sin x}{cos x} \Rightarrow 1 - cos x = sin x \Rightarrow cos x + sin x = 1 \Rightarrow \frac{1}{\sqrt{2}} cos x + \frac{1}{\sqrt{2}} sin x = \frac{1}{\sqrt{2}}$ $secx-1=tanxrArr1/cosx-1=sinx/cosxrArr1-cosx=sinxrArrcosx+sinx=1rArr1/sqrt(2)cosx+1/sqrt(2)sinx=1/sqrt(2)$
$\Rightarrow cos x cos (\frac{π}{4}) + sin x sin (\frac{π}{4}) = cos (\frac{π}{4})$ $rArrcosxcos(pi/4)+sinxsin(pi/4)=cos(pi/4)$
$\Rightarrow cos (x - \frac{π}{4}) = cos (\frac{π}{4})$ $rArrcos(x-pi/4)=cos(pi/4)$
$x - \frac{π}{4} = 2 k π \pm \frac{π}{4}, k \in Z$ $x-pi/4=2kpi+-pi/4,kinZ$
$x = 2 k π \pm \frac{π}{4} + \frac{π}{4}, k \in Z$ $x=2kpi+-pi/4+pi/4,kinZ$
$x = 2 k π + \frac{π}{4} + \frac{π}{4}, k \in Z or x = 2 k π - \frac{π}{4} + \frac{π}{4} . k \in Z$ $x=2kpi+pi/4+pi/4,kinZorx=2kpi-pi/4+pi/4.kinZ$
$x = 2 k π + \frac{π}{2}, k \in Z or x = 2 k π, k \in Z$ $color(red)(x=2kpi+pi/2,kinZorx=2kpi,kinZ)$
But, taking $k = 0, . . e t c .$ $k=0,..etc.$
$x = 2 k π + \frac{π}{2} . k \in Z \Rightarrow x = \frac{π}{2}$ $x=2kpi+pi/2.kinZ=>x=pi/2$ , which does not satisfy
$sec x - 1 = tan x$ $secx-1=tanx$ , as $sec (\frac{π}{2}) and tan (\frac{π}{2})$ $sec (pi/2) and tan(pi/2)$ are undefined.
So, $x \neq 2 k π + \frac{π}{2}, k \in Z \Rightarrow x = 2 k π, k \in Z$ $color(red)(x!=2kpi+pi/2,kinZrArrx=2kpi,kinZ)$

Answer 2 · 2018-03-06T14:22:36

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

Explanation:

$sec (x) - 1 = tan (x)$ $sec(x)−1=tan(x)$

$\frac{1}{cos x} - 1 = \frac{sin x}{cos x}$ $1/cosx −1=sinx/cosx$

$\frac{1 - sin x}{cos x} = 1$ $(1-sinx)/cosx =1$

$sin x + cos x = 1, where cos x \neq 0$ $sinx + cosx = 1, "where" cosx ≠ 0$

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

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