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受欢迎的三角函数 >

证明 ((1+cot(a)-csc(a))(1+tan(a)+sec(a)))=2

解答

证明

((1 + cot (a) - csc (a)) (1 + tan (a) + sec (a))) = 2

解答

真

求解步骤

(1 + cot (a) - csc (a)) (1 + tan (a) + sec (a)) = 2

调整左侧

(1 + cot (a) - csc (a)) (1 + tan (a) + sec (a))

用 sin, cos 表示

(1 + cot (a) - csc (a)) (1 + sec (a) + tan (a))

使用基本三角恒等式:

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

= (1 + \frac{cos ( a )}{sin ( a )} - csc (a)) (1 + sec (a) + tan (a))

使用基本三角恒等式:

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

= (1 + \frac{cos ( a )}{sin ( a )} - \frac{1}{sin ( a )}) (1 + sec (a) + tan (a))

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

(1 + \frac{cos ( a )}{sin ( a )} - \frac{1}{sin ( a )}) (1 + \frac{1}{cos ( a )} + \frac{sin ( a )}{cos ( a )}) : \frac{( sin ( a ) + cos ( a ) - 1 ) ( cos ( a ) + 1 + sin ( a ) )}{sin ( a ) cos ( a )}

(1 + \frac{cos ( a )}{sin ( a )} - \frac{1}{sin ( a )}) (1 + \frac{1}{cos ( a )} + \frac{sin ( a )}{cos ( a )})

化简

1 + \frac{cos ( a )}{sin ( a )} - \frac{1}{sin ( a )} : \frac{cos ( a ) - 1}{sin ( a )} + 1

1 + \frac{cos ( a )}{sin ( a )} - \frac{1}{sin ( a )}

合并分式

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

使用法则

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

= \frac{cos ( a ) - 1}{sin ( a )}

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

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

化简

1 + \frac{cos ( a ) - 1}{sin ( a )} : \frac{sin ( a ) + cos ( a ) - 1}{sin ( a )}

1 + \frac{cos ( a ) - 1}{sin ( a )}

将项转换为分式:

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

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

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 ( a ) + cos ( a ) - 1}{sin ( a )}

乘以:

1 \cdot sin (a) = sin (a)

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

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

合并分式

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

使用法则

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

= \frac{1 + sin ( a )}{cos ( a )}

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

化简

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

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

将项转换为分式:

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

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

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 ( a ) + 1 + sin ( a )}{cos ( a )}

乘以:

1 \cdot cos (a) = cos (a)

= \frac{cos ( a ) + 1 + sin ( a )}{cos ( a )}

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

分式相乘:

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

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

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

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

乘开

(- 1 + cos (a) + sin (a)) (1 + cos (a) + sin (a)) : cos^{2} (a) + 2 sin (a) cos (a) + sin^{2} (a) - 1

(- 1 + cos (a) + sin (a)) (1 + cos (a) + sin (a))

打开括号

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

使用加减运算法则

+ (- a) = - a

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

化简

- 1 \cdot 1 - 1 \cdot cos (a) - 1 \cdot sin (a) + 1 \cdot cos (a) + cos (a) cos (a) + cos (a) sin (a) + 1 \cdot sin (a) + sin (a) cos (a) + sin (a) sin (a) : cos^{2} (a) + 2 sin (a) cos (a) + sin^{2} (a) - 1

- 1 \cdot 1 - 1 \cdot cos (a) - 1 \cdot sin (a) + 1 \cdot cos (a) + cos (a) cos (a) + cos (a) sin (a) + 1 \cdot sin (a) + sin (a) cos (a) + sin (a) sin (a)

对同类项分组

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

同类项相加:

cos (a) sin (a) + sin (a) cos (a) = 2 sin (a) cos (a)

= - 1 \cdot cos (a) - 1 \cdot sin (a) + 1 \cdot cos (a) + cos (a) cos (a) + 2 sin (a) cos (a) + 1 \cdot sin (a) + sin (a) sin (a) - 1 \cdot 1

同类项相加:

- 1 \cdot cos (a) + 1 \cdot cos (a) = 0

= - 1 \cdot sin (a) + cos (a) cos (a) + 2 sin (a) cos (a) + 1 \cdot sin (a) + sin (a) sin (a) - 1 \cdot 1

同类项相加:

- 1 \cdot sin (a) + 1 \cdot sin (a) = 0

= cos (a) cos (a) + 2 sin (a) cos (a) + sin (a) sin (a) - 1 \cdot 1

cos (a) cos (a) = cos^{2} (a)

cos (a) cos (a)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (a) cos (a) = cos^{1 + 1} (a)

= cos^{1 + 1} (a)

数字相加:

1 + 1 = 2

= cos^{2} (a)

sin (a) sin (a) = sin^{2} (a)

sin (a) sin (a)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

sin (a) sin (a) = sin^{1 + 1} (a)

= sin^{1 + 1} (a)

数字相加:

1 + 1 = 2

= sin^{2} (a)

1 \cdot 1 = 1

1 \cdot 1

数字相乘:

1 \cdot 1 = 1

= 1

= cos^{2} (a) + 2 sin (a) cos (a) + sin^{2} (a) - 1

= cos^{2} (a) + 2 sin (a) cos (a) + sin^{2} (a) - 1

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

- 1 + 1 = 0

= \frac{2 cos ( a ) sin ( a )}{cos ( a ) sin ( a )}

约分:

cos (a)

= \frac{2 sin ( a )}{sin ( a )}

约分:

sin (a)

= 2

= 2

= 2

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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