Question 1078827
Solve 

{{{tan(x)+1=sec(x)}}}
<pre><b>
We square both sides

{{{tan^2(x)+2tan(x)+1=sec^2(x)}}}

We use the Pythagorean trig identity: 
{{{1+tan^2(theta)=sec^2(theta)}}} to write
the right side:

{{{tan^2(x)+2tan(x)+1=1+tan^2(x)}}}

{{{2tan(x)=0}}}

{{{tan(x)=0}}}

You finish. That gives you two values for x in [0,2<font face="symbol">p</font>).  
But since I squared both sides, you must check for 
extraneous solutions. 

--------------------------

{{{2tan(x)sec(x)-tan(x)=0}}}

Factor out common factor of tan(x)

{{{tan(x)(2sec(x)-1^"")=0}}}

{{{matrix(3,5,

tan(x)=0,";","","",2sec(x)-1=0,
"","","","",2sec(x)=1,
"","","","",sec(x)=1/2))}}}

You finish by solving tan(x)=0 for x.
But remember that sec(x) always has
absolute value greater than or equal to 1.

Edwin</pre></b>