Question 630556
<pre>
tan(x) + sec(x) = 1

{{{sin(x)/cos(x)}}} + {{{1/cos(x)}}} = 1

Multiply through by LCD of cos(x)

sin(x) + 1 = cos(x)

sin(x) - cos(x) = -1

Since sin({{{pi/4}}}) = cos({{{pi/4)}}}) = {{{sqrt(2)/2}}}

We can use that fact to make the left side into the the form of
the right side of the identity {{{sin(alpha-beta)=sin(alpha)cos(beta)-cos(alpha)sin(beta)}}}

We multiply through by {{{sqrt(2)/2}}}

sin(x){{{sqrt(2)/2}}} - cos(x){{{sqrt(2)/2}}} = -1{{{sqrt(2)/2}}}

Write the first {{{sqrt(2)/2}}} in the first term as cos({{{pi/4)}}}) and
the {{{sqrt(2)/2}}} in the second term as sin({{{pi/4}}}):

sin(x)cos({{{pi/4}}}) - cos(x)sin({{{pi/4}}}) = {{{-sqrt(2)/2}}}

Using the identity {{{sin(alpha-beta)=sin(alpha)cos(beta)-cos(alpha)sin(beta)}}}, we can rewrite the left side as

sin(x-{{{pi/4}}}) = {{{-sqrt(2)/2}}}

Therefore x-{{{pi/4}}} must be a 3rd or 4th quadrant angle to have
a negative value for its sine.   

Since 0 &#8806; x < 2{{{pi}}} we subtract {{{pi/4}}} from all three sides:

    0-{{{pi/4}}} &#8806; x-2{{{pi}}} < 2pi-{{{pi/4}}} 
      
    {{{-pi/4}}} &#8806; x-2{{{pi}}} < {{{7pi/4}}} 

The only angle in that interval which has a sine of {{{-sqrt(2)/2}}} 
is 4th quadrant angle {{{-pi/4}}} 
       

x-{{{pi/4}}} = {{{-pi/4}}}        

Solving for x:

   x = {{{-pi/4}}} + {{{pi/4}}}

   x = 0

It checks in the original (sometimes there are extraneous answers)


tan(x) + sec(x) = 1
tan(0) + sec(0) = 1
          0 + 1 = 1   

x = 0 is the only solution!

Edwin</pre>