25. Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for any real number $x$.對於實數 $x$, 求 $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ 的最小値。
25. $2 \sqrt{2}-1$
25. Let $a=\sin x$ and $b=\cos x$. Then $a^{2}+b^{2}=1$ and we want to minimise
$$
\left|a+b+\frac{a}{b}+\frac{b}{a}+\frac{1}{b}+\frac{1}{a}\right|=\left|a+b+\frac{1+a+b}{a b}\right| .
$$

If we set $c=a+b$, then $a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2}=\frac{c^{2}-1}{2}$ and so the above quantity becomes
$$
\left|c+\frac{2(1+c)}{c^{2}-1}\right|=\left|c+\frac{2}{c-1}\right|=\left|(c-1)+\frac{2}{c-1}+1\right| \text {. }
$$

Let $f(c)$ denote this quantity. As $c=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right), c$ may take values between $-\sqrt{2}$ and $\sqrt{2}$. On the other hand, for positive real number $r$ we have $r+\frac{2}{r}=\left(\sqrt{r}-\sqrt{\frac{2}{r}}\right)^{2}+2 \sqrt{2} \geq 2 \sqrt{2}$, with equality when $\sqrt{r}=\sqrt{\frac{2}{r}}$, i.e. $r=\sqrt{2}$. Hence
- if $c>1$, then $f(c) \geq 2 \sqrt{2}+1$;
- if $c<1$, then $f(c)=\left|1-\left[(1-c)+\frac{2}{1-c}\right]\right| \geq|1-2 \sqrt{2}|=2 \sqrt{2}-1$.

Equality in the second case holds if $1-c=\sqrt{2}$, i.e. if $c=1-\sqrt{2}$, which is possible since $-\sqrt{2}<1-\sqrt{2}<\sqrt{2}$. It follows that the answer is $2 \sqrt{2}-1$.