异曲同工：$1$ $+$ $\tan^{2} \alpha$ 与 $(\tan \alpha)^{\prime}$

一、前言

在数学中，通过寻找不同的公式之间的相同点或者差异点，可以让我们对公式的记忆与理解更加深入，例如：

$$
1 + \tan^{2} \alpha = \textcolor{orange}{\frac{1}{\cos ^{2} \alpha}}
$$

$$
(\tan \alpha)^{\prime} = \textcolor{orange}{\frac{1}{\cos ^{2} \alpha}}
$$

即：

$$
1 + \tan^{2} \alpha \textcolor{red}{=} (\tan \alpha)^{\prime}
$$

二、正文

下面是对 $1$ $+$ $\tan^{2} \alpha$ 与 $(\tan \alpha)^{\prime}$ 的推导计算过程：

$$
1 + \tan^{2} \alpha =
$$

$$
1 + \Big( \frac{\sin \alpha}{\cos \alpha} \Big)^{2} =
$$

$$
\frac{\cos ^{2} \alpha}{\cos ^{2} \alpha} + \frac{\sin ^{2} \alpha}{\cos ^{2} \alpha} =
$$

$$
\frac{\sin ^{2} \alpha + \cos ^{2} \alpha}{\cos ^{2} \alpha} =
$$

$$
\frac{1}{\cos ^{2} \alpha} = \sec ^{2} \alpha \Rightarrow
$$

$$
\textcolor{orange}{1 + \tan^{2} \alpha} = \textcolor{orange}{\frac{1}{\cos ^{2} \alpha}} = \textcolor{orange}{\sec ^{2} \alpha}.
$$

---

$$
(\tan \alpha)^{\prime} =
$$

$$
\Big( \frac{\sin \alpha}{\cos \alpha} \Big)^{\prime} =
$$

$$
\frac{(\sin \alpha)^{\prime} \cdot \cos \alpha \textcolor{red}{-} \sin \alpha \cdot (\cos \alpha)^{\prime}}{\cos ^{2} \alpha} =
$$

$$
\frac{\cos ^{2} \alpha \textcolor{red}{+} \sin ^{2} \alpha}{\cos ^{2} \alpha} =
$$

$$
\frac{1}{\cos ^{2} \alpha} = \sec ^{2} \alpha \Rightarrow
$$

$$
\textcolor{orange}{(\tan \alpha)^{\prime}} = \textcolor{orange}{\frac{1}{\cos ^{2} \alpha}} = \textcolor{orange}{\sec ^{2} \alpha}.
$$

---