How to prove that $\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$ How to prove this trigonometric identities ?
$$\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$$
Thank you.
 A: COMMENT.-In attention to the note by Community above I show the following:
$$LHS=\frac{\sin(\frac{15\pi}{18})}{\sin(\frac{11\pi}{18})\sin(\frac{4\pi}{18})}\\$$ Hence one has $$\sin(\frac{15\pi}{18})=4\cos(\frac{4\pi}{18})\sin(\frac{11\pi}{18})\sin(\frac{\pi}{18})$$ Note that $\dfrac{15\pi}{18}=150^{\circ}=180^{\circ}-30^{\circ}$ so one has  $$\sin(\frac{15\pi}{18})=\frac 12$$ Besides $$2\cos(\frac{4\pi}{18})\sin(\frac{11\pi}{18})=\sin(\frac{15\pi}{18})+\sin(\frac{7\pi}{18})$$ so we have $$\frac12=2\left(\frac12+\sin(\frac{7\pi}{18})\right)\sin(\frac{\pi}{18})$$ and because of $\frac{7\pi}{18}=90^{\circ}-20^{\circ}$ and $\frac{\pi}{18}=10^{\circ}$ we can reduce all to elementary calculations.
A: We have to check:
$$ -\tan\frac{\pi}{9}+\cot\frac{2\pi}{9}=4\cos\frac{4\pi}{9}\cot\frac{2\pi}{9} $$
hence, by multiplying both sides by $\sin\frac{2\pi}{9}=2\sin\frac{\pi}{9}\cos\frac{\pi}{9}$:
$$ -2\sin^2\frac{\pi}{9}+\cos\frac{2\pi}{9} = 4\cos\frac{4\pi}{9}\cos\frac{2\pi}{9} $$
so we just have to check that $x=\cos\frac{2\pi}{9}$ is a solution of:
$$ 2x-1 = 4x(2x^2-1) $$
or a root of:
$$ p(x) = 8x^3-6x+1 = 2\cdot T_3(x)-1, \tag{1}$$
where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $\cos\frac{2\pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(\cos\alpha)=\cos(3\alpha)$:
$$ 2\cdot T_3\left(\cos\frac{2\pi}{9}\right)-1 = 2\cdot\cos\frac{2\pi}{3}-1 = 0, $$
hence we're done.
A: $\cot\dfrac{11\pi}{18}=\cot\left(\pi-\dfrac{7\pi}{18}\right)=-\cot\dfrac{7\pi}{18}$
Now $\cot\dfrac{2\pi}9-\cot\dfrac{7\pi}{18}=\dfrac{\sin\left(\dfrac{7\pi}{18}-\dfrac{2\pi}9\right)}{\sin\dfrac{2\pi}9\cdot\sin\dfrac{7\pi}{18}}=\dfrac1{2\sin\dfrac{2\pi}9\cos\dfrac\pi9}$
( as $\sin\dfrac{7\pi}{18}=\cos\left(\dfrac\pi2-\dfrac{7\pi}{18}\right)=\cos\dfrac\pi9$)
which needs to be $=\sin\dfrac\pi{18}\cot\dfrac{2\pi}9$
which will be true if $8\sin\dfrac\pi{18}\cos\dfrac\pi9\cos\dfrac{2\pi}9=1$
$\iff8\cos\dfrac\pi9\cos\dfrac{2\pi}9\cos\dfrac{4\pi}9=1$
as $\dfrac{4\pi}9+\dfrac\pi{18}=\dfrac\pi2$
Now use: Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
