I'll assume this is:
Show
[tex]\dfrac{\sin 2a + \cos 2a}{2 \cos a + \sin a - 2(\cos ^3 a + \sin ^3 a)} = \csc a [/tex]
What a mess. Who comes up with these?
Somehow the whole thing has to factor so the numerator and denominator cancel and we're left with sin a in the denominator.
We have the sum of cubes so we'll probably need to know the factorization
[tex]x^3 + y^3 = (x + y) (x^2 - x y + y^2)[/tex]
Let's just start with the denominator and factor this way.
[tex]2 \cos a + \sin a - 2(\cos ^3 a + \sin ^3 a)[/tex]
[tex]= 2 \cos a + \sin a - 2(\cos a + \sin a)(\cos^2 a - \cos a \sin a + \sin ^2 a)[/tex]
[tex]= 2 \cos a + \sin a - 2(\cos a + \sin a)(1 - \cos a \sin a)[/tex]
[tex]= 2 \cos a + \sin a - 2\cos a-2 \sin a +2(\cos a + \sin a)\cos a \sin a[/tex]
[tex]= -\sin a +2(\cos a + \sin a)\cos a \sin a[/tex]
[tex]= \sin a (-1 + 2 \cos a(\cos a + \sin a))[/tex]
[tex]= \sin a(2 \cos^2 a - 1 + 2 \sin a \cos a)[/tex]
We recognize the double angle formulas,
[tex]= \sin a(\cos 2a + \sin 2a)[/tex]
So
[tex]\dfrac{\sin 2a + \cos 2a}{2 \cos a + \sin a - 2(\cos ^3 a + \sin ^3 a)}[/tex]
[tex]=\dfrac{\sin 2a + \cos 2a}{ \sin a(\cos 2a + \sin 2a)}[/tex]
[tex]=\dfrac{1}{\sin a}[/tex]
[tex]=\csc a \quad\checkmark[/tex]
Phew.
