1. Find the value of $\cot \left(\cot ^{-1} 2+\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5\right)$.
Answer: $\frac{5}{14}$
Solution: Using the identity
$$
\cot ^{-1} x+\cot ^{-1} y=\cot ^{-1}\left(\frac{x y-1}{x+y}\right),
$$
the expression $\cot ^{-1} 2+\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5$ can be simplified to $\cot ^{-1} \frac{9}{7}+\cot ^{-1} \frac{11}{7}$. Thus, we have
$$
\cot \left(\cot ^{-1} 2+\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5\right)=\cot \left(\cot ^{-1} \frac{9}{7}+\cot ^{-1} \frac{11}{7}\right)=\frac{5}{14}
$$