1. Let's start by stating the problem: We want to understand why the graphs of $\sec x$ and $\csc x$ are different even though they seem similar. 2. Recall the definitions: - $\sec x = \frac{1}{\cos x}$ - $\csc x = \frac{1}{\sin x}$ 3. Important rules: - The domain of $\sec x$ excludes points where $\cos x = 0$ because division by zero is undefined. - The domain of $\csc x$ excludes points where $\sin x = 0$ for the same reason. 4. Let's analyze the zeros of $\cos x$ and $\sin x$: - $\cos x = 0$ at $x = \frac{\pi}{2} + k\pi$, where $k$ is any integer. - $\sin x = 0$ at $x = k\pi$, where $k$ is any integer. 5. This means the vertical asymptotes (where the function goes to infinity) for $\sec x$ occur at $x = \frac{\pi}{2} + k\pi$, and for $\csc x$ at $x = k\pi$. 6. Because these asymptotes occur at different points, the graphs look different. 7. Also, the shape of $\cos x$ and $\sin x$ differ in phase (shifted by $\frac{\pi}{2}$), so their reciprocals reflect this difference. 8. In summary, although $\sec x$ and $\csc x$ are both reciprocals of trigonometric functions, their domains and asymptotes differ due to the zeros of $\cos x$ and $\sin x$, causing their graphs to be different. Final answer: The graphs of $\sec x$ and $\csc x$ differ because their vertical asymptotes occur at different points, reflecting the zeros of $\cos x$ and $\sin x$ respectively, and their phase difference causes distinct graph shapes.