# INTERVAL 0 2pi*sec x + tan x = 1

rakesh05 | Certified Educator

Given interval is `[0,2pi]` .

And given identity is   `secx+tanx=1`

which we can write as    `1/(cosx)+(sinx)/(cosx)=1`

`or` ,                             `(1+sinx)/(cosx)=1`

or,                             `(1+sinx)=cosx`

or,                            `1+sinx-cosx=0`

or,                          `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`

or,   `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`

or,        `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`

or,         `2sin(x/2)[cos(x/2)+sin(x/2)]=0`

`or,`    `2sin(x/2)=0`    and    `cos(x/2)+sin(x/2)=0` .

Now,     `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` ,  n=0,1,2.    As per the need of our problem.

`rArr x/2=npi`    or,     `x=2npi`  for n=0,1 as our interval is `[0,2pi]` .

So, `x=0`   or,  `x=2pi` .

Now, if    `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`

`rArrtan(x/2)=-1`     or,  `tan(x/2)=tan(pi/2+pi/4)`

or,           `x/2=(pi/2+pi/4)`

or,            `x=pi+pi/2`

or,            `x=3pi/2` .

Clearly,   `x=0`  and `x=2pi`  are the solutions satisfying our given identity over the given interval.

oldnick | Student

`secx+tanx=1`

since :  `cosx=1/sqrt(1+tan^2x)`       `secx= sqrt(1+tan^2x)`

subsituing:

`sqrt(1+tan^2x)+tanx=1`

`sqrt(1+tan^2x)=1-tanx`

squaring both sides:

`1+tan^2x=1-2tanx+tan^2x`

So:

`2tanx=0`    `x= kpi`

pramodpandey | Student

`sec(x)+tan(x)=1`

`1/cos(x)+sin(x)/cos(x)=1`

`(1+sin(x))/cos(x)=1`

`(sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2))/(cos^2(x/2)-sin^2(x/2))=1`

`(cos(x/2)+sin(x/2))^2/{(cos(x/2)+sin(x/2))(cos(x/2)-sin(x/2))}=1`

`(cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2))=1`

`(cos(x/2)(1+sin(x/2)/cos(x/2)))/(cos(x/2)(1-sin(x/2)/cos(x/2)))=1`

`(1+tan(x/2))/(1-tan(x/2))=1`

`(tan(pi/4)+tan(x/2))/(1-tan(pi/4)tan(x/2))=1`

`tan(pi/4+x/2)=1`

`tan(pi/4+x/2)=tan(pi/4)`

`pi/4+x/2=npi+pi/4` , n is an integer.

`if n=0 =>x=0`

`if n=1=>x=2pi`