<s>[INST] How to prove that $\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1$ From this geometry problem, I can not find geometry solution.
However the answer is $X=\frac{2\pi}{15}$ by geometry method.

Then I get the identity $$\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1.$$
How to prove it by trigonometric method ?
Thank in advances.
 [/INST] This is not a fully sketched answer.I enjoyed while going through it and hence thought others might like it. 
Since $\displaystyle\sin\frac{\pi}{10}=\frac{1}{4}(\sqrt{5}-1)$ we obtain $\displaystyle\sin\frac{\pi}{30}=\frac{1}{8}(\sqrt{30-6\sqrt{5}}-1-\sqrt{5})$ and consequently $$\tan\frac{\pi}{30}=\sqrt{7-2\sqrt{5}-2\sqrt{15-6\sqrt{5}}}$$
Now since $\displaystyle\cos\frac{\pi}{5}=\frac{1}{4}(\sqrt{5}+1)$ we obtain $\displaystyle\cos\frac{\pi}{15}=\frac{1}{8}(\sqrt{30+6\sqrt{5}}-1+\sqrt{5})$ and consequently $$\tan\frac{\pi}{15}=\frac{1}{\sqrt{7+2\sqrt{5}+2\sqrt{15+6\sqrt{5}}}}$$ which in turn implies $$\tan\frac{2\pi}{15}=\frac{\sqrt{7+2\sqrt{5}+2\sqrt{15+6\sqrt{5}}}}{3+\sqrt{5}+\sqrt{15+6\sqrt{5}}}$$ 
Finally note that $\displaystyle\cos\frac{\pi}{5}=\frac{1}{4}(\sqrt{5}+1)$. Putting all these together and after some good bit of time spent on simplification we obtain the result.
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