Question

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

Answer 1

`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

`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"`