INTERVAL 0 2pi*sec x + tan x = 1 

Expert Answers
rakesh05 eNotes educator| 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


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





squaring both sides:



`2tanx=0`    `x= kpi`

pramodpandey | Student












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

`if n=0 =>x=0`

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

so answer is