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受欢迎的三角函数 >

证明 (1+tan(x))/(1-tan(x))+(1+cot(x))/(1-cot(x))=0

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

= \frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 - \frac{c o s ( x )}{s i n ( x )}} + \frac{1 + tan ( x )}{1 - tan ( x )}

使用基本三角恒等式:

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

= \frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 - \frac{c o s ( x )}{s i n ( x )}} + \frac{1 + \frac{s i n ( x )}{c o s ( x )}}{1 - \frac{s i n ( x )}{c o s ( x )}}

\frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 - \frac{c o s ( x )}{s i n ( x )}} + \frac{1 + \frac{s i n ( x )}{c o s ( x )}}{1 - \frac{s i n ( x )}{c o s ( x )}} = 0

\frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 - \frac{c o s ( x )}{s i n ( x )}} + \frac{1 + \frac{s i n ( x )}{c o s ( x )}}{1 - \frac{s i n ( x )}{c o s ( x )}}

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

\frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 - \frac{c o s ( x )}{s i n ( x )}}

化简

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

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

将项转换为分式:

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

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

乘以:

1 \cdot sin (x) = sin (x)

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

= \frac{1 + \frac{c o s ( x )}{s i n ( x )}}{\frac{s i n ( x ) - c o s ( x )}{s i n ( x )}}

化简

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

1 + \frac{cos ( x )}{sin ( x )}

将项转换为分式:

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

= \frac{1 \cdot sin ( x )}{sin ( x )} + \frac{cos ( x )}{sin ( 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 sin ( x ) + cos ( x )}{sin ( x )}

乘以:

1 \cdot sin (x) = sin (x)

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

= \frac{\frac{s i n ( x ) + c o s ( x )}{s i n ( x )}}{\frac{s i n ( x ) - c o s ( x )}{s i n ( x )}}

分式相除:

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

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

约分:

sin (x)

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

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

\frac{1 + \frac{s i n ( x )}{c o s ( x )}}{1 - \frac{s i n ( x )}{c o s ( x )}}

化简

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

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

将项转换为分式:

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

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

乘以:

1 \cdot cos (x) = cos (x)

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

= \frac{1 + \frac{s i n ( x )}{c o s ( x )}}{\frac{c o s ( x ) - s i n ( x )}{c o s ( x )}}

化简

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

1 + \frac{sin ( x )}{cos ( x )}

将项转换为分式:

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

= \frac{1 \cdot cos ( x )}{cos ( x )} + \frac{sin ( x )}{cos ( 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 ( x ) + sin ( x )}{cos ( x )}

乘以:

1 \cdot cos (x) = cos (x)

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

= \frac{\frac{c o s ( x ) + s i n ( x )}{c o s ( x )}}{\frac{c o s ( x ) - s i n ( x )}{c o s ( x )}}

分式相除:

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

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

约分:

cos (x)

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

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

sin (x) - cos (x), cos (x) - sin (x)

的最小公倍数:

- (sin (x) - cos (x))

sin (x) - cos (x), cos (x) - sin (x)

最小公倍数 (LCM)

cos (x) - sin (x) = - (sin (x) - cos (x))

= sin (x) - cos (x), - (sin (x) - cos (x))

计算出由出现在

sin (x) - cos (x)

或

cos (x) - sin (x)

中的因子组成的表达式

= - (sin (x) - cos (x))

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

- (sin (x) - cos (x))

对于

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

将分母和分子乘以

- 1

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

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

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

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

消掉

\frac{- ( sin ( x ) + cos ( x ) ) + cos ( x ) + sin ( x )}{( sin ( x ) - cos ( x ) )} : 0

\frac{- ( sin ( x ) + cos ( x ) ) + cos ( x ) + sin ( x )}{( sin ( x ) - cos ( x ) )}

分解

- (sin (x) + cos (x)) + cos (x) + sin (x) : 0

- (sin (x) + cos (x)) + cos (x) + sin (x)

改写为

= - 1 \cdot (cos (x) + sin (x)) + 1 \cdot (cos (x) + sin (x))

因式分解出通项

(cos (x) + sin (x))

= (cos (x) + sin (x)) (- 1 + 1)

整理后得

= 0

= \frac{0}{( sin ( x ) - cos ( x ) )}

整理后得

= 0

= - 0

= 0

= 0

= 0

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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