1. **Problem Statement:** We analyze the cotangent function $\cot \alpha$ defined on the interval $(0^\circ, 90^\circ)$ using a right-angled triangle and its properties. 2. **Definition and Formula:** By definition, for $\alpha \in (0^\circ, 90^\circ)$, $$\cot \alpha = \frac{1}{\tan \alpha}.$$ Recall that $\tan \alpha = \frac{\text{opposite}}{\text{adjacent}}$ in a right triangle, so $\cot \alpha = \frac{\text{adjacent}}{\text{opposite}}$. 3. **Domain and Range:** - The domain is $(0^\circ, 90^\circ)$ because $\tan \alpha$ is defined and nonzero there. - As $\alpha \to 0^\circ$, $\tan \alpha \to 0$, so $\cot \alpha = \frac{1}{\tan \alpha} \to +\infty$. - As $\alpha \to 90^\circ$, $\tan \alpha \to +\infty$, so $\cot \alpha = \frac{1}{\tan \alpha} \to 0^+$. Thus, the range of $\cot \alpha$ is $(0, +\infty)$. 4. **Continuity and Undefined Points:** - $\cot \alpha$ is not defined at $\alpha = 90^\circ$ because $\tan 90^\circ$ is undefined (vertical asymptote). 5. **Boundedness:** - Since $\cot \alpha$ approaches infinity near $0^\circ$ and zero near $90^\circ$, it is **not bounded**. 6. **Monotonicity:** - The derivative of $\cot \alpha$ is $-\csc^2 \alpha$, which is negative for all $\alpha \in (0^\circ, 90^\circ)$. - Therefore, $\cot \alpha$ is a strictly decreasing function on this interval. **Summary of options:** - The range is $(0, +\infty)$: **True**. - The cotangent function is not defined for $\alpha = 90^\circ$: **True**. - The cotangent function is a bounded function: **False**. - The cotangent function is decreasing: **True**. **Final answer:** The correct statements are that the range is $(0, +\infty)$, $\cot \alpha$ is not defined at $90^\circ$, and $\cot \alpha$ is decreasing. It is not bounded.