Question 1208308
.
1/tan^2pie/8 - 1/sin^2pie/8
Evaluate
~~~~~~~~~~~~~~~~~



        I will evaluate an expression    {{{1/tan^2(x)}}} - {{{1/sin^2(x)}}}    for arbitrary  x =/= 0, +/-{{{pi}}}, +/-{{{2pi}}}, . . . 



<pre>
We have  {{{1/tan^2(x)}}}} = {{{cos^2(x)/sin^2(x)}}}.


THEREFORE,  {{{1/tan^2(x)}}} - {{{1/sin^2(x)}}} = {{{cos^2(x)/sin^2(x)}}} - {{{1/sin^2(x)}}} = {{{(cos^2(x)-1)/sin^2(x)}}}.


The numerator is  {{{-sin^2(x)}}}, by Pythagoras.  THEREFORE, the last fraction in the previous line is -1.


<U>ANSWER</U>.  The expression  {{{1/tan^2(x)}}} - {{{1/sin^2(x)}}}  is identically equal to -1 

         for all values of x different from 0, +/-{{{pi}}}, +/-{{{2pi}}}, . . . 

         In particular,  {{{1/tan^2(pi/8)}}} - {{{1/sin^2(pi/8)}}}  is -1.
</pre>

Solved for much more general expression, in simplest possible way, in 3 lines with complete explanations.