Question 1208297
<pre>

{{{(1/(tan(x) + cot(x) + sec(x) + csc(x)))^2}}} 

{{{1^""/(tan(x) + cot(x) + sec(x) + csc(x)^"")^2}}}

Draw a right triangle with angle x, hypotenuse 1, and legs "a" and "b"

{{{drawing(100,850/11,-.2, 4.2,-.2,3.2,
locate(.6,.7,x), red(arc(0,0,2.5,-2.5,0,37)),locate(1.54,2,1),
triangle(0,0,4,0,4,3), locate(2,.7,b), locate(3.6,1.8,a) )}}}
 
{{{1^""/(a/b + b/a + 1/b + 1/a)^2}}}

Since {{{a/b+b/a+1/b+1/a=(a^2+b^2+a+b)/(ab^"")=(1+a+b)/(ab)}}}, that's
because by the Pythagorean theorem, a<sup>2</sup>+b<sup>2</sup>=1

{{{1^""/((1+a+b)/(ab))^2}}}

Multiply top and bottom by a<sup>2</sup>b<sup>2</sup>

{{{a^2b^2/(1+a+b)^2}}}

We multiply out the denominator,

{{{a^2b^2/(1+a^2+b^2+2a+2b+2ab)}}}

Since a<sup>2</sup>+b<sup>2</sup>=1,

{{{a^2b^2/(2+2a+2b+2ab)}}}

{{{expr(1^""/2^"")*(a^2b^2/(1+a+b+ab^""))}}}

The denominator factors

{{{expr(1^""/2^"")*(a^2b^2/((1+a)(1+b)))}}}

Multiply by the product of the "conjugates" of the
factors on the bottom over itself. (Sort of like
rationalizing the denominator)


{{{expr(1^""/2^"")*(a^2b^2/((1+a)(1+b)))}}}{{{""*""}}}{{{(((1-a)(1-b)^"")/((1-a)(1-b)^""))}}}

 
{{{expr(1/2)*(a^2b^2(1-a)(1-b))/((1-a^2)(1-b^2)))}}}

By the Pythagorean theorem  1-a<sup>2</sup>=b<sup>2</sup> and 1-b<sup>2</sup>=a<sup>2</sup>

{{{expr(1/2)*(a^2b^2(1-a)(1-b))/(b^2a^2)))}}}

Canceling:

{{{expr(1/2)*(1-a)(1-b)}}}

Since a = sin(x) and b = cos(x)

{{{expr(1/2)*(1^""-sin(x))(1^""-cos(x))}}}  <--simplest form

Edwin</pre>