It's not clear if this is a problem to solve or a problem to prove. Let's see where it goes.
We note the cotangent half angle formula is
[tex]\cot x = \dfrac{ 1 + \cos 2x}{\sin 2x}[/tex]
The tangent and cotangent half angle is expressible in terms of the full angle without any ambiguity, so let's set b=a/4 so a=4b.
It turns out to be true for all a (at least all a that don't make the any of the functions undefined). So it's a problem to prove.
Here's the proof. I actually did it from the bottom up, but it's better to present it this way as a proof.
We start with the cosine double angle formula:
[tex]\cos 2b = 2\cos^2 b- 1[/tex]
Multipy both sides by sin b:
[tex]\sin b \cos 2b = 2 \sin b \cos^2 b - \sin b[/tex]
Sine double angle formula:
[tex]\sin b \cos 2b = \sin 2b \cos b- \sin b[/tex]
Add sin 2b to both sides:
[tex] \sin 2b+ \sin b\cos 2b = \sin 2b + \sin 2b \cos b - \sin b[/tex]
Divide by sin b sin 2b
[tex] \dfrac{1}{\sin b} + \dfrac{\cos 2b}{\sin 2b} = \dfrac{1 + \cos b}{\sin b} - \dfrac{1}{\sin 2b}[/tex]
Turn to cosecants and cotangents. We use the cotangent half angle formula above.
[tex] \csc b + \cot 2b = \cot \frac b 2 - \csc 2b[/tex]
Substituting b=a/4:
[tex] \csc \frac a 4 + \cot \frac a 2 = \cot \frac a 8 - \csc \frac a 2 \quad\checkmark[/tex]
