Question 1208309
<pre>

Note: &pi; is spelled "pi", not "pie" like you eat. LOL
  
{{{1^""/tan^2(pi/8)-1^""/sin^2(pi/8)}}}

Two well known identities are

{{{tan(expr(1/2)x)}}}{{{""=""}}}{{{"" +- sqrt((1-cos(x))/(1+cos(x)))}}}
and
{{{sin(expr(1/2)x)}}}{{{""=""}}}{{{"" +- sqrt((1-cos(x))/(2))}}}

Squaring both 

{{{tan^2(expr(1/2)x)}}}{{{""=""}}}{{{(1-cos(x))/(1+cos(x))}}}
and
{{{sin^2(expr(1/2)x)}}}{{{""=""}}}{{{(1-cos(x))/2}}}

Substituting {{{pi/4}}}

{{{tan^2(pi/8)}}}{{{""=""}}}{{{(1-cos(pi/4))/(1+cos(pi/4))}}}
and
{{{sin^2(pi/8)}}}{{{""=""}}}{{{(1-cos(pi/4))/2^""}}}

Substituting {{{cos(pi/4)}}}{{{""=""}}}{{{sqrt(2)/2}}}

{{{tan^2(pi/8)}}}{{{""=""}}}{{{(1-sqrt(2)/2)/(1+sqrt(2)/2)}}}{{{""=""}}}{{{(2-sqrt(2))/(2+sqrt(2))}}}
and
{{{sin^2(pi/8)}}}{{{""=""}}}{{{(1-sqrt(2)/2)/2^""}}}{{{""=""}}}{{{(2-sqrt(2))/4}}}

reciprocating both:

{{{1^""/tan^2(pi/8)}}}{{{""=""}}}{{{(2+sqrt(2))/(2-sqrt(2))}}}
{{{1^""/sin^2(pi/8)}}}{{{""=""}}}{{{4/(2-sqrt(2))}}}
and
{{{1^""/tan^2(pi/8)-1^""/sin^2(pi/8)}}}{{{""=""}}}{{{(2+sqrt(2))/(2-sqrt(2))-4/(2-sqrt(2))}}}{{{""=""}}}{{{(-2+sqrt(2))/(2-sqrt(2))}}}{{{""=""}}}{{{(-(2-sqrt(2)))/(2-sqrt(2))}}}{{{""=""}}}{{{""=""}}}{{{(-(cross(2-sqrt(2))))/(cross(2-sqrt(2)))}}}{{{""=""}}}{{{-1}}}

Answer = -1

Edwin</pre>