Question 1006779
<pre>
{{{tan^2("157.5°")=tan^2("315°"/2)}}}

There are at least 6 well-known formulas for the tangent of 
half an angle.  Since 315° is a special angle, (a QIV angle
with a reference angle of 45°), we can use any of these:

{{{tan(theta/2)}}}{{{""=""}}}{{{sin(theta)/(1+cos(theta))}}}

{{{tan(theta/2)}}}{{{""=""}}}{{{tan(theta)/(1+sec(theta))}}}

{{{tan(theta/2)}}}{{{""=""}}}{{{1/(csc(theta)+cot(theta))}}}

{{{tan(theta/2)}}}{{{""=""}}}{{{(1+cos(theta))/sin(theta)}}}

{{{tan(theta/2)}}}{{{""=""}}}{{{(sec(theta)-1)/tan(theta))}}}

{{{tan(theta/2)}}}{{{""=""}}}{{{csc(theta)-cot(theta))}}}

The last one is the simplest as it has no denominator. The others
will require more complicated simplification, but will all end
up with the same expression:

{{{tan("315°"/2)}}}{{{""=""}}}{{{""=""}}}{{{csc("315°")-cot("315°"))}}}{{{""=""}}}{{{-sqrt(2)-(-1))}}}{{{""=""}}}{{{1-sqrt(2)}}}

Edwin</pre>