May Long
01/18/2024 · High School
x) emostar qua: \( \operatorname{aec}^{4} \theta-\sec ^{2} \theta=\frac{1}{\cot ^{4} \theta}+\frac{1}{\cot ^{2} \theta} \)
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为了证明等式 \( \operatorname{sec}^{4} \theta - \sec^{2} \theta = \frac{1}{\cot^{4} \theta} + \frac{1}{\cot^{2} \theta} \),我们可以按照以下步骤进行:
### 步骤 1:将等式左边转换为基本三角函数形式
首先,将等式左边转换为基本三角函数形式:
\[ \operatorname{sec}^{4} \theta - \operatorname{sec}^{2} \theta \]
我们知道,\(\operatorname{sec} \theta = \frac{1}{\cos \theta}\),所以:
\[ \operatorname{sec}^{4} \theta = \left(\frac{1}{\cos \theta}\right)^{4} = \frac{1}{\cos^{4} \theta} \]
\[ \operatorname{sec}^{2} \theta = \left(\frac{1}{\cos \theta}\right)^{2} = \frac{1}{\cos^{2} \theta} \]
因此,等式左边可以写为:
\[ \operatorname{sec}^{4} \theta - \operatorname{sec}^{2} \theta = \frac{1}{\cos^{4} \theta} - \frac{1}{\cos^{2} \theta} \]
### 步骤 2:将等式右边转换为基本三角函数形式
接下来,将等式右边转换为基本三角函数形式:
\[ \frac{1}{\cot^{4} \theta} + \frac{1}{\cot^{2} \theta} \]
我们知道,\(\cot \theta = \frac{\cos \theta}{\sin \theta}\),所以:
\[ \cot^{4} \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^{4} = \frac{\cos^{4} \theta}{\sin^{4} \theta} \]
\[ \cot^{2} \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^{2} = \frac{\cos^{2} \theta}{\sin^{2} \theta} \]
因此,等式右边可以写为:
\[ \frac{1}{\cot^{4} \theta} + \frac{1}{\cot^{2} \theta} = \frac{\sin^{4} \theta}{\cos^{4} \theta} + \frac{\sin^{2} \theta}{\cos^{2} \theta} \]
### 步骤 3:化简等式左边
现在,我们化简等式左边:
\[ \frac{1}{\cos^{4} \theta} - \frac{1}{\cos^{2} \theta} \]
将两个分数合并:
\[ \frac{1}{\cos^{4} \theta} - \frac{1}{\cos^{2} \theta} = \frac{\cos^{2} \theta - 1}{\cos^{4} \theta} \]
我们知道,\(\cos^{2} \theta - 1 = -\sin^{2} \theta\),所以:
\[ \frac{\cos^{2} \theta - 1}{\cos^{4} \theta} = \frac{-\sin^{2} \theta}{\cos^{4} \theta} \]
### 步骤 4:化简等式右边
现在,我们化简等式右边:
\[ \frac{\sin^{4} \theta}{\cos^{4} \theta} + \frac{\sin^{2} \theta}{\cos^{2} \theta} \]
将两个分数合并:
\[ \frac{\sin^{4} \theta}{\cos^{4} \theta} + \frac{\sin^{2} \theta}{\cos^{2} \theta} = \frac{\sin^{2} \theta (\sin^{2} \theta + 1)}{\cos^{4} \theta} \]
我们知道,\(\sin^{2} \theta + 1 = \cos^{2} \theta\),所以:
\[ \frac{\sin^{2} \theta (\sin^{2} \theta + 1)}{\cos^{4} \theta} = \frac{\sin^{2} \theta \cos^{2} \theta}{\cos^{4} \theta} = \frac{\sin^{2} \theta}{\cos^{2} \theta} \]
### 步骤 5:比较等式两边
现在,我们比较等式两边:
等式左边:
\[ \frac{-\sin^{2} \theta}{\cos^{4} \theta} \]
等式右边:
\[ \frac{\sin^{
Quick Answer
等式左边和右边相等,因此等式成立。
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