Hello again.
I'm asked to find the exact value of tan 15deg.
To make it short, I got (3 - sqrt(3)) / (3 + sqrt(3)),
but in the back of the book, they got 2 - sqrt(2).
How am I suppose to get that from what I got?
Thanks.
$\displaystyle \tan{(60-45)} = \frac{\tan{60}-\tan{45}}{1+\tan{60}\tan{45}} = \frac{\sqrt{3}-1}{1+(\sqrt{3})(1)}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
From here it's just rationalizing the denominator:
$\displaystyle =\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{4-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$