9. If $\tan x+\tan y=5$ and $\tan (x+y)=10$, find $\cot ^{2} x+\cot ^{2} y$.
Answer: 96
Solution: We know that
$$
\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=10
$$
This leads to
$$
\tan x \tan y=\frac{1}{2} .
$$

Hence,
$$
\begin{aligned}
\cot ^{2} x+\cot ^{2} y & =\frac{1}{\tan ^{2} x}+\frac{1}{\tan ^{2} y} \\
& =\frac{\tan ^{2} y+\tan ^{2} x}{\tan ^{2} x \tan ^{2} y} \\
& =\frac{(\tan x+\tan y)^{2}-2 \tan x \tan y}{(\tan x \tan y)^{2}} \\
& =\frac{5^{2}-1}{\frac{1}{4}} \\
& =96 .
\end{aligned}
$$