Can you prove this trigonometric identity?

[anemone]

Prove that $\dfrac{1+\sin 6^{\circ}+\cos 12^{\circ}}{\cos 6^{\circ}+\sin 12^{\circ}}=\sqrt{3} $.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. laura123
2. kaliprasad

I'll show both solutions since they are different in the method of approaching.(Yes)

Here's laura123's solution:

Spoiler

Since $\cos 36°=\dfrac{\sqrt{5}+1}{4}$ and $\sin 18°=\dfrac{\sqrt{5}-1}{4}$ it follows:

$\cos 36°-\sin 18°=\dfrac{1}{2}$
$\cos (30°+6°)-\sin (30°-12°)=\dfrac{1}{2}$
$\cos 30°\cos 6°-\sin 30°\sin 6°-\sin 30°\cos 12°+\cos 30°\sin 12°=\dfrac{1}{2}$
$\dfrac{\sqrt{3}}{2}\cos 6°-\dfrac{1}{2}\sin 6°-\dfrac{1}{2}\cos 12°+\dfrac{\sqrt{3}}{2}\sin 12°=\dfrac{1}{2}$
$\sqrt{3}\cos 6°-\sin 6°-\cos 12°+\sqrt{3}\sin 12°=1$
$\sqrt{3}\cos 6°+\sqrt{3}\sin 12°=1+\sin 6°+\cos 12°$
$\sqrt{3}(\cos 6°+\sin 12°)=1+\sin 6°+\cos 12°$
$\dfrac{\sqrt{3}(\cos 6°+\sin 12°)}{\cos 6°+\sin 12°}=\dfrac{1+\sin 6°+\cos 12°}{\cos 6°+\sin 12°}$
$\sqrt{3}=\dfrac{1+\sin 6°+\cos 12°}{\cos 6°+\sin 12°}$.
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Here's kaliprasad's solution:

Spoiler

First, note that

$\cos(36^\circ) - \cos(72^\circ)$
= $2 * \sin(18^\circ) * \sin(54^\circ)$
= $\dfrac{1}{2} * \dfrac{\sin(2*18)^\circ * \sin(2*54)^\circ }{\cos(18)^\circ * \cos(54)^\circ}$
= $\dfrac{1}{2} * \dfrac{\sin(36^\circ) * \sin(108^\circ)}{ (\sin(108^\circ) * \sin(36^\circ)})$
= $\dfrac{1}{2}$

We then have
$1+ \sin\ 6^\circ + \cos \ 12^\circ$
= $2(\dfrac{1}{2} +\dfrac{1}{2} \sin\ 6^\circ + \dfrac{1}{2}\cos \ 12^\circ)$
= $2(\dfrac{1}{2} +\cos \ 60^\circ \sin\ 6^\circ + \cos \ 60^\circ\cos \ 12^\circ)$
= $2 ( \dfrac{1}{2}- \sin (60-6)^\circ + \sin 60^\circ \cos 6^\circ + \cos 72^\circ + \sin 60^\circ \sin 12^\circ )$
= $2 ( \dfrac{1}{2}- \sin\ 54^\circ + \sin 60^\circ \cos 6^\circ + \cos 72^\circ + \sin 60^\circ \sin 12^\circ )$
= $2 ( \sin 60^\circ (\cos 6^\circ + \sin 12^\circ ) + \dfrac{1}{2} + \cos 72^\circ - \cos 36^\circ )$
=$2 ( \sin 60^\circ (\cos 6^\circ + \sin 12^\circ ) + \dfrac{1}{2} -( \cos 36^\circ-\cos 72^\circ )$
=$2 ( \sin 60^\circ (\cos 6^\circ + \sin 12^\circ ) + \dfrac{1}{2} -\dfrac{1}{2}$
= $2 \sin 60^\circ ( \cos 6^\circ + \sin 12^\circ )$
= $\dfrac{ 1 + \sin 6^\circ + \cos 12^\circ }{\cos 6^\circ + \sin 12^\circ } = 2 \sin 60^\circ = \sqrt(3)$