24. Find the least possible value of $f(x)=\frac{9}{1+\cos 2 x}+\frac{25}{1-\cos 2 x}$, where $x$ ranges over all real numbers for which $f(x)$ is defined.
24. Answer: 32
For all real values of $x$ for which $f(x)$ is defined, we have
$$
\begin{aligned}
f(x) & =\frac{9}{1+\cos 2 x}+\frac{25}{1-\cos 2 x}=\frac{9}{2 \cos ^{2} x}+\frac{25}{2 \sin ^{2} x} \\
& =\frac{1}{2}\left(9 \tan ^{2} x+9\right)+\frac{1}{2}\left(25 \cot ^{2} x+25\right) \\
& =17+\frac{1}{2}\left(9 \tan ^{2} x+25 \cot ^{2} x\right) \\
& \geq 17+\frac{1}{2}\left(2 \sqrt{\left(9 \tan ^{2} x\right)\left(25 \cot ^{2} x\right)}\right. \text { (AM-GM Inequality) } \\
& =17+\frac{1}{2}(2 \sqrt{9 \times 25}) \\
& =32 .
\end{aligned}
$$

Note that $f\left(\tan ^{-1} \sqrt{\frac{5}{3}}\right)=32$. Thus the least possible value of $f(x)$ is 32 .