Question 276355
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First let's learn where the Pythagorean identities came from,
namely the Pythagorean theorem:

{{{drawing(266+2/3,400,-1,1,-1,2, triangle(-.5,0,.5,0,.5,1.6), 
rectangle(.4,0,.5,.1), locate(-.35,.17,theta), locate(-.2,0,adjacent),
locate(.55,.8,opposite), locate(-.75,.8,hypotenuse) )}}}

By the Pythagorean theorem:

{{{adjacent^2 + opposite^2 = hypotenuse^2}}}

Divide through by {{{hypotenuse^2}}}

{{{(adjacent^2)/(hypotenuse^2) + (opposite^2)/(hypotenuse^2) = (hypotenuse^2)/(hypotenuse^2)}}}

{{{((adjacent)/(hypotenuse))^2 + ((opposite)/(hypotenuse))^2 = 1}}}

{{{Cos^2theta + Sin^2theta = 1}}}

That's the first Pythagorean identity:

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{{{adjacent^2 + opposite^2 = hypotenuse^2}}}

Divide through by {{{adjacent^2}}}

{{{(adjacent^2)/(adjacent^2) + (opposite^2)/(adjacent^2) = (hypotenuse^2)/(adjacent^2)}}}

{{{1 + ((opposite)/(hypotenuse))^2 = ((hypotenuse)/adjacent))^2}}}

{{{1 + Tan^2theta = Sec^2theta}}}

That's the second Pythagorean identity:

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{{{adjacent^2 + opposite^2 = hypotenuse^2}}}

Divide through by {{{opposite^2}}}

{{{(adjacent^2)/(opposite^2) + (opposite^2)/(opposite^2) = (hypotenuse^2)/(opposite^2)}}}

{{{((adjacent)/(opposite))^2 + 1 = ((hypotenuse)/(opposite))^2}}}

{{{Cot^2theta + 1 = Csc^2theta}}}

That's the third Pythagorean identity:

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Be sure to memorize all three of them.  Usually we write them
this way, so they're easier to learn:

{{{Sin^2theta + Cos^2theta = 1}}}
{{{1 + Tan^2theta = Sec^2theta}}}
{{{1 + Cot^2theta = Csc^2theta}}}

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Those identities are true for ALL angles {{{theta}}}

So for

{{{Csc^2}}}{{{(3alpha)}}}{{{""-""}}}{{{Cot^2}}}{{{(3alpha)}}}

Since the third Pythagorean identity

{{{1 + Cot^2theta = Csc^2theta}}} 

is true for ALL angles {{{theta}}}, let {{{theta = 3alpha}}}, then

{{{1 + Cot^2}}}{{{(3alpha)}}}{{{""=""}}}{{{Csc^2}}}{{{(3alpha)}}}

So we can replace {{{Csc^2}}}{{{(3alpha)}}} by {{{1 + Cot^2}}}{{{(3alpha)}}}

So your problem:

{{{Csc^2}}}{{{(3alpha)}}}{{{""-""}}}{{{Cot^2}}}{{{(3alpha)}}}

becomes:

{{{1 + Cot^2}}}{{{(3alpha)}}}{{{""-""}}}{{{Cot^2}}}{{{(3alpha)}}}

then the "cotangent squared"s cancel:

{{{1 + cross(Cot^2)}}}{{{(cross(3alpha))}}}{{{""-""}}}{{{cross(Cot^2)}}}{{{(cross(3alpha))}}}

and that just leaves 1, which is an integer.

Edwin</pre>