A very effective way to prove most trig identities like the one above is to use complex numbers. To do this we use the formula

$e^{ix}=\cos (x)+i\sin (x)$

So $\cos (x)=(1/2)(e^{ix}+e^{-ix})$

and $\sin (x)=(1/2i)(e^{ix}-e^{-ix})$

So if I set $x=i\pi /11$, then the question above is to show that:

${\displaystyle \frac{x^4-x^{-4}}{x^4+x^{-4}}+2(x^1-x^{-1})=i\sqrt{11}}$

Now if we multiply both sides by $(x^4+x^{-4})$ and then square each side and simplify then we get (after a little work which I'm certainly not typing out!):

$4(x^{10}+x^9+x^8-x^7+x^6+2x^2+1+x^{-10}+x^{-9}+x^{-8}-x^{-7}+x^{-6}+2x^{-2})=0$

Now $x^{11}=-1$ (because $x^{11}=e^{i\pi }=-1$)

So $-1=x^{1}1=x^{9+2}=x^9\times x^2\Rightarrow x^9+x^2=0$

And similarly $-x^7=x^{-4}$

So the long equation above becomes:

$(x^{10}+x^8+x^6+x^4+x^2+x^{-10}+x^{-8}+x^{-6}+x^{-4}+x^{-2}+1)=0$

So if we can prove this is true then the result you want will be true.

Now, $x^{11}=-1$ means that $x^{22}=1$

So $0=x^{22}-1=(x-1)(1+x+x^2+x^3+\dots +x^{21})$

$=(x-1)(1+x)(1+x^2+x^4+x^6+\dots +x^{20})$

Now, clearly $x$ isn't $1$ or $-1$, so the above means that:

$1+x^2+x^4+\dots +x^{20}=0$

Dividing by $x^{10}$ (allowed as $x$ isn't 0) gives:

$x^{10}+x^8+x^6+x^4+x^2+x^{-10}+x^{-8}+x^{-6}+x^{-4}+x^{-2}+1=0$

Which is exactly what we wanted.