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受欢迎的三角函数 >

证明 tan^2(y)(csc^2(y)-1)=1

解答

证明

tan^{2} (y) (csc^{2} (y) - 1) = 1

解答

真

求解步骤

tan^{2} (y) (csc^{2} (y) - 1) = 1

调整左侧

tan^{2} (y) (csc^{2} (y) - 1)

用 sin, cos 表示

(- 1 + csc^{2} (y)) tan^{2} (y)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= (- 1 + (\frac{1}{sin ( y )})^{2}) tan^{2} (y)

使用基本三角恒等式:

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

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

化简

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

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

(\frac{1}{sin ( y )})^{2} = \frac{1}{sin ^{2} ( y )}

(\frac{1}{sin ( y )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( y )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( y )}

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

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

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

使用指数法则:

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

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

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

分式相乘:

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

= \frac{sin ^{2} ( y ) ( - 1 + \frac{1}{s i n ^{2} ( y )} )}{cos ^{2} ( y )}

化简

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

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

将项转换为分式:

1 = \frac{1 sin ^{2} ( y )}{sin ^{2} ( y )}

= - \frac{1 \cdot sin ^{2} ( y )}{sin ^{2} ( y )} + \frac{1}{sin ^{2} ( y )}

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 sin ^{2} ( y ) + 1}{sin ^{2} ( y )}

乘以:

1 \cdot sin^{2} (y) = sin^{2} (y)

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

= \frac{\frac{- s i n ^{2} ( y ) + 1}{s i n ^{2} ( y )} sin ^{2} ( y )}{cos ^{2} ( y )}

乘

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

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

分式相乘:

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

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

约分:

sin^{2} (y)

= - - sin^{2} (y) + 1

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= \frac{cos ^{2} ( y )}{cos ^{2} ( y )}

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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