Question

Coseca/4+cota/2=cota/8-coseca/2​

Answer 1

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

$\cot x = \dfrac{ 1 + \cos 2x}{\sin 2x}$

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:

$\cos 2b = 2\cos^2 b- 1$

Multipy both sides by sin b:

$\sin b \cos 2b = 2 \sin b \cos^2 b - \sin b$

Sine double angle formula:

$\sin b \cos 2b = \sin 2b \cos b- \sin b$

Add sin 2b to both sides:

$\sin 2b+ \sin b\cos 2b = \sin 2b + \sin 2b \cos b - \sin b$

Divide by sin b sin 2b

$\dfrac{1}{\sin b} + \dfrac{\cos 2b}{\sin 2b} = \dfrac{1 + \cos b}{\sin b} - \dfrac{1}{\sin 2b}$

Turn to cosecants and cotangents.  We use the cotangent half angle formula above.

$\csc b + \cot 2b = \cot \frac b 2 - \csc 2b$

Substituting b=a/4:

$\csc \frac a 4 + \cot \frac a 2 = \cot \frac a 8 - \csc \frac a 2 \quad\checkmark$