Question

Prove that $\tan 821/2=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)$

Answer 1

$\tan 30=\frac{1}{\sqrt{3}}$

$\frac{2\tan 15}{1-{\tan }^{2}15}=\frac{1}{\sqrt{3}}$ Let $\tan 15=a$

$2\sqrt{3}a=1-a^2$

$a^2+2\sqrt{3}a-1=0$

$a=\frac{-2\sqrt{3}\pm \sqrt{12+4}}{2}(\because \tan 15>0)$

$\therefore \tan 15=2-\sqrt{3}$

$\frac{2\tan 7\frac{1}{2}°}{1-{\tan }^{2}7\frac{1}{2}°}=2-\sqrt{3}$ Let $\tan 7\frac{1}{2}°=b$

$2b=(2-\sqrt{3})(1-b^2)$

$(2-\sqrt{3})b^2+2b-(2-\sqrt{3})=0$

$\therefore b=\frac{-2+\sqrt{4+4{(2-\sqrt{3})}^2}}{2(2-\sqrt{3})}(\because \tan 7\frac{1}{2}°>0)$

$=\frac{-1+\sqrt{8-4\sqrt{3}}}{(2-\sqrt{3})}$

$\therefore \tan 7\frac{1}{2}°=\frac{\sqrt{6}-\sqrt{2}-1}{2-\sqrt{3}}$

$\tan 82\frac{1}{2}°=\cot 7\frac{1}{2}°=\frac{2-\sqrt{3}}{\sqrt{6}-\sqrt{2}-1}$

$=\frac{(2-\sqrt{3})(\sqrt{6}+\sqrt{2}+1)}{(\sqrt{6}-(\sqrt{2}+1))(\sqrt{6}+\sqrt{2}+1)}$

$=\frac{2\sqrt{6}-3\sqrt{2}+2\sqrt{2}-\sqrt{6}+2-\sqrt{3}}{6-{(\sqrt{2}+1)}^2}$

$=\frac{\sqrt{6}-\sqrt{2}+2-\sqrt{3}}{3-2\sqrt{2}}$

$=\frac{(\sqrt{2}-1)(\sqrt{3}+\sqrt{2})}{{(\sqrt{2}-1)}^2}$

$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}-1}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)$

$\therefore \tan 82\frac{1}{2}°=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)$