Question 1209582
<pre>
I think we can get by just using the standard trig identities that we're
all accustomed to using, even though it might take a little longer:

{{{cot^2(alpha)cot^2(beta)}}}{{{""=""}}}{{{3}}}

{{{expr(cos^2(alpha)/sin^2(alpha))*expr(cos^2(beta)/sin^2(alpha)))}}}{{{""=""}}}{{{3}}}

{{{expr(  (1-sin^2(alpha))/sin^2(alpha)  )*expr(  (1-sin^2(beta))/sin^2(beta)  )}}}{{{""=""}}}{{{3}}}

{{{(1-sin^2(alpha)-sin^2(beta)+sin^2(alpha)sin^2(beta))/(sin^2(alpha)sin^2(beta))}}}{{{""=""}}}{{{3}}}

{{{1-sin^2(alpha)-sin^2(beta)+sin^2(alpha)sin^2(beta)}}}{{{""=""}}}{{{3(sin^2(alpha)sin^2(beta)^"")}}}

{{{1-sin^2(alpha)-sin^2(beta)}}}{{{""=""}}}{{{2(sin^2(alpha)sin^2(beta)^"")}}}

{{{1}}}{{{""=""}}}{{{sin^2(alpha)+ 2sin^2(alpha)sin^2(beta)+sin^2(beta)}}} <-- call this equation 1

Now let's work on

{{{(2-cos(2alpha)^"")(2-cos(2beta)^"")}}}

{{{(2-(1-2sin^2(alpha)^"")^"")(2-(1-2sin^2(beta)^"")^"")}}}

{{{(1+2sin^2(alpha)^"")(1+2sin^2(beta)^"")}}}

{{{1+2sin^2(alpha)+2sin^2(beta)+4sin^2(alpha)sin^2(beta)}}}

{{{1+2(sin^2(alpha)+sin^2(beta)+2sin^2(alpha)sin^2(beta)^"")}}}

Now look up at equation 1 and you see that what's in the big parentheses
here is equal to 1, so,

{{{1+2(1)}}}{{{""=""}}}{{{3}}}

So the answer is 3.

Edwin</pre>