Showing that $\frac{ 1- \sin\frac{5\pi}{18}}{\sqrt{3} \sin \frac{5\pi}{18}}= \tan\frac{\pi}{18} $ It's easy to verify on WolframAlpha, but I have difficulty deriving it. 
It's easy to see 
$$
\tan\left(\frac{\pi}{18}\right)=\frac{\sqrt{3}-\tan(5/18 \pi)}{1+\sqrt{3}\tan(5/18 \pi)}
$$ 
 A: We need to prove that $$(1-\sin50^{\circ})\cos10^{\circ}=2\sin50^{\circ}\sin10^{\circ}\cos30^{\circ}$$ or
$$2\cos10^{\circ}-\sin60^{\circ}-\sin40^{\circ}=2(\cos40^{\circ}-\cos60^{\circ})\cos30^{\circ}$$ or
$$2\cos10^{\circ}-\sin40^{\circ}=2\cos40^{\circ}\cos30^{\circ}$$ or
$$2\cos10^{\circ}-\sin40^{\circ}=\cos70^{\circ}+\cos10^{\circ}$$ or
$$\cos10^{\circ}=\cos70^{\circ}+\cos50^{\circ}.$$
Can you end it now?
A: Evaluate,
$$\frac1{\sin\frac{5\pi}{18}}-\sqrt3 \tan\frac{\pi}{18}
=\frac{\cos\frac{\pi}{18}-\sqrt3 \sin\frac{5\pi}{18}\sin\frac{\pi}{18}}{\sin\frac{5\pi}{18}\cos\frac{\pi}{18}}\tag 1$$
Examine the numerator,
$$\cos\frac{\pi}{18}-\frac{\sqrt3}2\cdot2 \sin\frac{5\pi}{18}\sin\frac{\pi}{18}$$
$$= \cos\frac{\pi}{18}-\cos\frac\pi6(\cos\frac{2\pi}{9}-\cos\frac{\pi}{3})$$
$$= \cos\frac{\pi}{18}-\frac12(\cos\frac{\pi}{18} + \cos\frac{7\pi}{18})+\cos\frac{\pi}{6}\cos\frac{\pi}{3}$$
$$= \frac12(\cos\frac{\pi}{18} - \cos\frac{7\pi}{18})+\frac12\cos\frac{\pi}{6}$$
$$= \sin\frac{\pi}{6} \sin\frac{2\pi}{9}+\frac12\cos\frac{\pi}{6}$$
$$= \frac12( \sin\frac{2\pi}{9}+\sin\frac{\pi}{3})
=\sin\frac{5\pi}{18}\cos\frac{\pi}{18}$$
Substitute the result for the numerator into (1) to have
$$\frac1{\sin\frac{5\pi}{18}}-\sqrt3 \tan\frac{\pi}{18}= 1$$
Then, rearrange to obtain,
$$\frac{ 1- \sin\frac{5\pi}{18}}{\sqrt{3} \sin\frac{5\pi}{18}}= \tan\frac{\pi}{18}$$
A: Hint 1: The double angle formula for $\tan$ is $$\tan\left(a-b\right)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$
Hint 2: For the equation with $\sin$, use the identity $\tan x=\dfrac{\sin x}{\cos x}$.
A: Use the formula for $\tan(x-y)$:
$$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$$
You also need to know that $\tan\frac{\pi}{3}=\sqrt{3}$
A: Hint 1: Express $\frac \pi {18}$ as a sum (difference) of 2 other angles.
Hint 2: For proving questions, I usually find that starting with the simpler side of the equation and working towards the other is easier. In this case, I would work from RHS to LHS.
Hint 3: Use the addition (subtraction) formula for $\tan$.
