SOLUTION: Solve the equation for exact solutions over the interval [0,2{{{pi}}}). tan(x) + sec(x) = 1

tan(x) + sec(x) = 1

 +  = 1

Multiply through by LCD of cos(x)

sin(x) + 1 = cos(x)

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

Since sin() = cos() = 

We can use that fact to make the left side into the the form of
the right side of the identity 

We multiply through by 

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

Write the first  in the first term as cos() and
the  in the second term as sin():

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

Using the identity , we can rewrite the left side as

sin(x-) = 

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

Since 0 ≦ x < 2 we subtract  from all three sides:

    0- ≦ x-2 < 2pi- 
      
     ≦ x-2 <  

The only angle in that interval which has a sine of  
is 4th quadrant angle  
       

x- =         

Solving for x:

   x =  + 

   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