Question 972445
<pre>
{{{1^""/sin(2A^"")+cos(4A^"")/sin(4A^"")}}}{{{""=""}}}{{{cot(A^"")-csc(4A^"")}}}

First we work with the left side:

{{{1^""/sin(2A^"")+(2cos^2(2A)-1)/(2sin(2A^"")cos(2A^""))}}}
Get LCD
{{{2cos(2A^"")/(2sin(2A^"")cos(2A^""))+(2cos^2(2A)-1)/(2sin(2A^"")cos(2A^""))}}}

{{{(2cos(2A^"")+2cos^2(2A)-1)/(2sin(2A^"")cos(2A^""))}}}

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Now we work with the right side:

{{{cot(A^"")-csc(4A^"")}}}

There is an identity {{{tan(x/2)}}}{{{""=""}}}{{{sin(x^"")/(1+cos(x^""))}}}
We can flip both sides and get

{{{cot(x/2)}}}{{{""=""}}}{{{(1+cos(x^""))/sin(x^"")}}}

If we let {{{A=x/2}}} and therefore {{{x=2A}}}, we get

{{{cot(A^"")}}}{{{""=""}}}{{{(1+cos(2A^""))/sin(2A^"")}}}

So the right side of the identity that we are to prove becomes:

{{{(1+cos(2A^""))/sin(2A^"")-csc(4A^"")}}}

{{{(1+cos(2A^""))/sin(2A^"")-1^""/sin(4A^"")}}}

{{{(1+cos(2A^""))/sin(2A^"")-1^""/(2sin(2A^"")cos(2A^""))}}}

Get LCD:

{{{((1+cos(2A^""))(2cos(2A^"")))/(2sin(2A^"")cos(2A^""))-1^""/(2sin(2A^"")cos(2A^""))}}}

{{{(2cos(2A^"")+2cos^2(2A)-1)/(2sin(2A^"")cos(2A^""))}}}

We have worked each side separately to the same expression.
So we have proved the identity.

Edwin</pre>