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受欢迎的三角函数 >

sec^2(x)-1= 1/(cot(x))

解答

sec^{2} (x) - 1 = \frac{1}{cot ( x )}

解答

x = \frac{π}{4} + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) - 1 = \frac{1}{cot ( x )}

两边进行平方

(sec^{2} (x) - 1)^{2} = (\frac{1}{cot ( x )})^{2}

两边减去

(\frac{1}{cot ( x )})^{2}

(sec^{2} (x) - 1)^{2} - \frac{1}{cot ^{2} ( x )} = 0

化简

(sec^{2} (x) - 1)^{2} - \frac{1}{cot ^{2} ( x )} : \frac{cot ^{2} ( x ) ( sec ^{2} ( x ) - 1 ) ^{2} - 1}{cot ^{2} ( x )}

(sec^{2} (x) - 1)^{2} - \frac{1}{cot ^{2} ( x )}

将项转换为分式:

(sec^{2} (x) - 1)^{2} = \frac{( sec ^{2} ( x ) - 1 ) ^{2} cot ^{2} ( x )}{cot ^{2} ( x )}

= \frac{( sec ^{2} ( x ) - 1 ) ^{2} cot ^{2} ( x )}{cot ^{2} ( x )} - \frac{1}{cot ^{2} ( x )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{( sec ^{2} ( x ) - 1 ) ^{2} cot ^{2} ( x ) - 1}{cot ^{2} ( x )}

\frac{cot ^{2} ( x ) ( sec ^{2} ( x ) - 1 ) ^{2} - 1}{cot ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cot^{2} (x) (sec^{2} (x) - 1)^{2} - 1 = 0

使用三角恒等式改写

- 1 + (- 1 + sec^{2} (x))^{2} cot^{2} (x)

使用毕达哥拉斯恒等式:

sec^{2} (x) = tan^{2} (x) + 1

sec^{2} (x) - 1 = tan^{2} (x)

= - 1 + (tan^{2} (x))^{2} cot^{2} (x)

(tan^{2} (x))^{2} = tan^{4} (x)

(tan^{2} (x))^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= tan^{2 \cdot 2} (x)

数字相乘:

2 \cdot 2 = 4

= tan^{4} (x)

= - 1 + tan^{4} (x) cot^{2} (x)

- 1 + cot^{2} (x) tan^{4} (x) = 0

分解

- 1 + cot^{2} (x) tan^{4} (x) : (tan^{2} (x) cot (x) + 1) (tan^{2} (x) cot (x) - 1)

- 1 + cot^{2} (x) tan^{4} (x)

将

- 1 + cot^{2} (x) tan^{4} (x)

改写为

- 1 + (cot (x) tan^{2} (x))^{2}

- 1 + cot^{2} (x) tan^{4} (x)

使用指数法则:

a^{b c} = (a^{b})^{c}

tan^{4} (x) = (tan^{2} (x))^{2}

= - 1 + cot^{2} (x) (tan^{2} (x))^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

cot^{2} (x) (tan^{2} (x))^{2} = (cot (x) tan^{2} (x))^{2}

= - 1 + (cot (x) tan^{2} (x))^{2}

= - 1 + (cot (x) tan^{2} (x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

- 1 + (cot (x) tan^{2} (x))^{2} = (cot (x) tan^{2} (x) + 1) (cot (x) tan^{2} (x) - 1)

= (cot (x) tan^{2} (x) + 1) (cot (x) tan^{2} (x) - 1)

(tan^{2} (x) cot (x) + 1) (tan^{2} (x) cot (x) - 1) = 0

分别求解每个部分

tan^{2} (x) cot (x) + 1 = 0 or tan^{2} (x) cot (x) - 1 = 0

tan^{2} (x) cot (x) + 1 = 0 : x = \frac{3 π}{4} + πn

tan^{2} (x) cot (x) + 1 = 0

使用三角恒等式改写

1 + cot (x) tan^{2} (x)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= 1 + cot (x) (\frac{1}{cot ( x )})^{2}

cot (x) (\frac{1}{cot ( x )})^{2} = \frac{1}{cot ( x )}

cot (x) (\frac{1}{cot ( x )})^{2}

(\frac{1}{cot ( x )})^{2} = \frac{1}{cot ^{2} ( x )}

(\frac{1}{cot ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cot ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( x )}

= \frac{1}{cot ^{2} ( x )} cot (x)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot cot ( x )}{cot ^{2} ( x )}

乘以:

1 \cdot cot (x) = cot (x)

= \frac{cot ( x )}{cot ^{2} ( x )}

约分:

cot (x)

= \frac{1}{cot ( x )}

= 1 + \frac{1}{cot ( x )}

1 + \frac{1}{cot ( x )} = 0

在两边乘以

cot (x)

1 + \frac{1}{cot ( x )} = 0

在两边乘以

cot (x)

1 \cdot cot (x) + \frac{1}{cot ( x )} cot (x) = 0 \cdot cot (x)

化简

1 \cdot cot (x) + \frac{1}{cot ( x )} cot (x) = 0 \cdot cot (x)

化简

1 \cdot cot (x) : cot (x)

1 \cdot cot (x)

乘以:

1 \cdot cot (x) = cot (x)

= cot (x)

化简

\frac{1}{cot ( x )} cot (x) : 1

\frac{1}{cot ( x )} cot (x)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot cot ( x )}{cot ( x )}

约分:

cot (x)

= 1

化简

0 \cdot cot (x) : 0

0 \cdot cot (x)

使用法则

0 \cdot a = 0

= 0

cot (x) + 1 = 0

cot (x) + 1 = 0

cot (x) + 1 = 0

将

1

到右边

cot (x) + 1 = 0

两边减去

1

cot (x) + 1 - 1 = 0 - 1

化简

cot (x) = - 1

cot (x) = - 1

验证解

找到无定义的点（奇点）:

cot (x) = 0

取

1 + \frac{1}{cot ( x )}

的分母，令其等于零

cot (x) = 0

以下点无定义

cot (x) = 0

将不在定义域的点与解相综合:

cot (x) = - 1

cot (x) = - 1

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

tan^{2} (x) cot (x) - 1 = 0 : x = \frac{π}{4} + πn

tan^{2} (x) cot (x) - 1 = 0

使用三角恒等式改写

- 1 + cot (x) tan^{2} (x)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 1 + cot (x) (\frac{1}{cot ( x )})^{2}

cot (x) (\frac{1}{cot ( x )})^{2} = \frac{1}{cot ( x )}

cot (x) (\frac{1}{cot ( x )})^{2}

(\frac{1}{cot ( x )})^{2} = \frac{1}{cot ^{2} ( x )}

(\frac{1}{cot ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cot ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( x )}

= \frac{1}{cot ^{2} ( x )} cot (x)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot cot ( x )}{cot ^{2} ( x )}

乘以:

1 \cdot cot (x) = cot (x)

= \frac{cot ( x )}{cot ^{2} ( x )}

约分:

cot (x)

= \frac{1}{cot ( x )}

= - 1 + \frac{1}{cot ( x )}

- 1 + \frac{1}{cot ( x )} = 0

在两边乘以

cot (x)

- 1 + \frac{1}{cot ( x )} = 0

在两边乘以

cot (x)

- 1 \cdot cot (x) + \frac{1}{cot ( x )} cot (x) = 0 \cdot cot (x)

化简

- 1 \cdot cot (x) + \frac{1}{cot ( x )} cot (x) = 0 \cdot cot (x)

化简

- 1 \cdot cot (x) : - cot (x)

- 1 \cdot cot (x)

乘以:

1 \cdot cot (x) = cot (x)

= - cot (x)

化简

\frac{1}{cot ( x )} cot (x) : 1

\frac{1}{cot ( x )} cot (x)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot cot ( x )}{cot ( x )}

约分:

cot (x)

= 1

化简

0 \cdot cot (x) : 0

0 \cdot cot (x)

使用法则

0 \cdot a = 0

= 0

- cot (x) + 1 = 0

- cot (x) + 1 = 0

- cot (x) + 1 = 0

将

1

到右边

- cot (x) + 1 = 0

两边减去

1

- cot (x) + 1 - 1 = 0 - 1

化简

- cot (x) = - 1

- cot (x) = - 1

两边除以

- 1

- cot (x) = - 1

两边除以

- 1

\frac{- cot ( x )}{- 1} = \frac{- 1}{- 1}

化简

cot (x) = 1

cot (x) = 1

验证解

找到无定义的点（奇点）:

cot (x) = 0

取

- 1 + \frac{1}{cot ( x )}

的分母，令其等于零

cot (x) = 0

以下点无定义

cot (x) = 0

将不在定义域的点与解相综合:

cot (x) = 1

cot (x) = 1

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

将解代入原方程进行验证

将它们代入

sec^{2} (x) - 1 = \frac{1}{cot ( x )}

检验解是否符合

去除与方程不符的解。

检验

\frac{3 π}{4} + πn

的解:

假

\frac{3 π}{4} + πn

代入

n = 1

\frac{3 π}{4} + π 1

对于

sec^{2} (x) - 1 = \frac{1}{cot ( x )}

代入

x = \frac{3 π}{4} + π 1

sec^{2} (\frac{3 π}{4} + π 1) - 1 = \frac{1}{cot ( \frac{3 π}{4} + π 1 )}

整理后得

1 = - 1

\Rightarrow 假

检验

\frac{π}{4} + πn

的解:

真

\frac{π}{4} + πn

代入

n = 1

\frac{π}{4} + π 1

对于

sec^{2} (x) - 1 = \frac{1}{cot ( x )}

代入

x = \frac{π}{4} + π 1

sec^{2} (\frac{π}{4} + π 1) - 1 = \frac{1}{cot ( \frac{π}{4} + π 1 )}

整理后得

1 = 1

\Rightarrow 真

x = \frac{π}{4} + πn

作图

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