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

2 Answers
maganbhai P.
Mar 6, 2018

#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)#

mizoo
Mar 6, 2018

#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"#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
21520 views around the world
You can reuse this answer
Creative Commons License