Problem 1207

## Solution

We have $\cos ^{2} t+\sin ^{2} t=1$. Hence $\cos t=\cos t /\left(\sqrt{ }\left(\cos ^{2} t+\sin ^{2} t\right)\right)=1 /\left(\sqrt{ }\left(1+\tan ^{2} t\right)\right)$. So if we put $\tan t=\sqrt{x}$, then $\cos \tan ^{-1} \sqrt{x}=1 / \sqrt{ }(1+x)$. We also have $\cos (\pi / 2-x)=\sin x$ and $\tan (\pi / 2-x)=1 / \tan x$. So $\tan ^{-1} \cos ^{-1} \sin ^{-1} \tan ^{-}$ ${ }^{1} x=1 / x(1)$. Hence also $\tan ^{-1} \cos ^{-1} \sin ^{-1} \tan ^{-1} \cos \tan ^{-1} \sqrt{x}=\sqrt{ }(x+1)$ (2). These two relations solve the problem.

Using (2) and iterating we can get $\sqrt{ }$ for any positive integer $n$. Hence in particular we can get $\mathrm{n}$ for any positive integer $\mathrm{n}$. Now suppose we want $\mathrm{m} / \mathrm{n}$ with $\mathrm{m}$ and $\mathrm{n}$ relatively prime. We show that $\mathrm{m} / \mathrm{n}$ can be achieved by induction on $n$. We have just dealt with the case $\mathrm{n}=1$. Suppose we have dealt with all $\mathrm{a} / \mathrm{b}$ with $\mathrm{b}$ $\mathrm{n}$, then we can write $\mathrm{m}=\mathrm{qn}+\mathrm{r}$ with $0<\mathrm{r}<\mathrm{n}$ and use (2) to reduce the problem to getting $\mathrm{r} / \mathrm{n}$. If $\mathrm{m}<\mathrm{n}$, then put $\mathrm{r}=\mathrm{m}$. Now use (1) to reduce the problem to $\mathrm{n} / \mathrm{r}$, which is solved by induction.