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From $ \tan 2x=\frac{2\tan x}{1-\tan^{2}x}$ and $ \tan 30^{0}=\frac{1}{\sqrt{3}}$ we we'll find $ \tan 15^{0}$ and $ \tan 7^{0}30'$.

Prove that \[\tan 7^\circ 30^{\prime }=\sqrt{6}+\sqrt{2}-\sqrt{3}-2.\]

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What is the $x'$ notation?

it is subdividion of degree of angles $1 ^o =60 ' =3600''$ one degree $^o$ is divided into 60 minutes (of arc) $'$, one minute $'$ into 60 seconds (of arc) $ ''$

parmenides51 wrote: Prove that \[\tan 7^\circ 30^{\prime }=\sqrt{6}+\sqrt{2}-\sqrt{3}-2.\]] Assuming this, same as above works. Note that $tan 15^\circ = 2 - \sqrt{3}$, and solve for $tan 7^\circ 30^{\prime }$ by double angle formulae. Alternatively, if one only knows the definition of $tan$, one can simply use the angle bisector theorem twice on a $30-60-90$ triangle.