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受欢迎的三角函数 >

证明 (sec^2(x)-1)cot^2(x)=1

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

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

使用基本三角恒等式:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

化简

(- 1 + (\frac{1}{cos ( x )})^{2}) (\frac{cos ( x )}{sin ( x )})^{2} : \frac{- cos ^{2} ( x ) + 1}{sin ^{2} ( x )}

(- 1 + (\frac{1}{cos ( x )})^{2}) (\frac{cos ( x )}{sin ( x )})^{2}

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

= (\frac{cos ( x )}{sin ( x )})^{2} (\frac{1}{cos ^{2} ( x )} - 1)

(\frac{cos ( x )}{sin ( x )})^{2} = \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

(\frac{cos ( x )}{sin ( x )})^{2}

使用指数法则:

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

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )} (\frac{1}{cos ^{2} ( x )} - 1)

分式相乘:

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

= \frac{cos ^{2} ( x ) ( - 1 + \frac{1}{c o s ^{2} ( x )} )}{sin ^{2} ( x )}

化简

- 1 + \frac{1}{cos ^{2} ( x )} : \frac{- cos ^{2} ( x ) + 1}{cos ^{2} ( x )}

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

将项转换为分式:

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

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

乘以:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

= \frac{\frac{- c o s ^{2} ( x ) + 1}{c o s ^{2} ( x )} cos ^{2} ( x )}{sin ^{2} ( x )}

乘

cos^{2} (x) \frac{- cos ^{2} ( x ) + 1}{cos ^{2} ( x )} : - cos^{2} (x) + 1

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

分式相乘:

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

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

约分:

cos^{2} (x)

= - - cos^{2} (x) + 1

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

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

= \frac{1 - cos ^{2} ( x )}{sin ^{2} ( x )}

使用三角恒等式改写

\frac{1 - cos ^{2} ( x )}{sin ^{2} ( x )}

使用毕达哥拉斯恒等式:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

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

使用法则

\frac{a}{a} = 1

= 1

= 1

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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