\# 9 Let $h$ be the piece-wise function $h(x)=\left\{\begin{array}{l}(x-0.042) \sec x, x \leq 1 \\ 1+\tan x, \quad x>1\end{array}\right.$. Find the zero of the function in the interval $[-1,2]$.

Step 1 ：The function is piece-wise, so we need to consider two cases: \(x \leq 1\) and \(x > 1\).

Step 2 ：For \(x \leq 1\), the function is \((x-0.042) \sec x\). We can set this equal to zero and solve for \(x\).

Step 3 ：The solution for the first case is \(x = 0.042\). This is within the interval \([-1,2]\), so it is a valid solution.

Step 4 ：For \(x > 1\), the function is \(1+\tan x\). We can set this equal to zero and solve for \(x\).

Step 5 ：There are no solutions for the second case in the interval \((1,2]\).

Step 6 ：Therefore, the only zero of the function in the interval \([-1,2]\) is \(x = 0.042\).

Step 7 ：Final Answer: The zero of the function in the interval \([-1,2]\) is \(\boxed{0.042}\).