Question

Find the positive integer p, q, r, s satisfying $\tan \frac{\pi }{24}=\left(\sqrt{p}-\sqrt{q}\right)\left(\sqrt{r}-s\right).$

Answer 1

$\tan \frac{\pi }{24}=\tan \left(\frac{\frac{\pi }{12}}{2}\right)$

Let $\frac{\pi }{24}=\theta \Rightarrow \frac{\pi }{12}=2\theta \Rightarrow \frac{\pi }{6}=4\theta$

$\tan 4\theta =\frac{1}{\sqrt{3}}$

$\frac{2\tan \left(2\theta \right)}{1-{\tan }^2\left(2\theta \right)}=\frac{1}{\sqrt{3}}$

$2\sqrt{3}\tan \left(2\theta \right)=1-{\tan }^2\left(2\theta \right)$

${\tan }^2\left(2\theta \right)+2\sqrt{3}\tan \left(2\theta \right)-1=0$

$\tan 2\theta =\frac{-2\sqrt{3}+\sqrt{12+4}}{2}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$

$\therefore \frac{2\tan \theta }{1-{\tan }^2\theta }=2-\sqrt{3}$

$\Rightarrow (2-\sqrt{3})({\tan }^2\theta +2\tan \theta -(2-\sqrt{3})=0$

$\tan \theta =\frac{-2+\sqrt{4+4(2-\sqrt{3})(2-\sqrt{3})}}{2(2-\sqrt{3})}$

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

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

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

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

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

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

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

$\tan \left(\frac{\pi }{24}\right)=(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)$

Given, $\tan \left(\frac{\pi }{24}\right)=(\sqrt{p}-\sqrt{q})(\sqrt{r}-s)$

$\therefore p=3,q=2,r=2,s=1$