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受欢迎的三角函数 >

(tan^2(b)+1)/(tan(b))=csc^2(b)

解答

\frac{tan ^{2} ( b ) + 1}{tan ( b )} = csc^{2} (b)

解答

b = \frac{π}{4} + πn

+1

度数

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

求解步骤

\frac{tan ^{2} ( b ) + 1}{tan ( b )} = csc^{2} (b)

两边减去

csc^{2} (b)

\frac{tan ^{2} ( b ) + 1}{tan ( b )} - csc^{2} (b) = 0

化简

\frac{tan ^{2} ( b ) + 1}{tan ( b )} - csc^{2} (b) : \frac{tan ^{2} ( b ) + 1 - csc ^{2} ( b ) tan ( b )}{tan ( b )}

\frac{tan ^{2} ( b ) + 1}{tan ( b )} - csc^{2} (b)

将项转换为分式:

csc^{2} (b) = \frac{csc ^{2} ( b ) tan ( b )}{tan ( b )}

= \frac{tan ^{2} ( b ) + 1}{tan ( b )} - \frac{csc ^{2} ( b ) tan ( b )}{tan ( b )}

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

\frac{tan ^{2} ( b ) + 1 - csc ^{2} ( b ) tan ( b )}{tan ( b )} = 0

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

tan^{2} (b) + 1 - csc^{2} (b) tan (b) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

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

使用基本三角恒等式:

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

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

化简

1 + (\frac{1}{cot ( b )})^{2} - (1 + cot^{2} (b)) \frac{1}{cot ( b )} : 1 + \frac{1}{cot ^{2} ( b )} - \frac{1 + cot ^{2} ( b )}{cot ( b )}

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

(1 + cot^{2} (b)) \frac{1}{cot ( b )}

分式相乘:

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

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

1 \cdot (1 + cot^{2} (b)) = 1 + cot^{2} (b)

1 \cdot (1 + cot^{2} (b))

乘以:

1 \cdot (1 + cot^{2} (b)) = (1 + cot^{2} (b))

= (1 + cot^{2} (b))

去除括号:

(a) = a

= 1 + cot^{2} (b)

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

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

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

1 - \frac{1 + cot ^{2} ( b )}{cot ( b )} + \frac{1}{cot ^{2} ( b )} = 0

用替代法求解

1 - \frac{1 + cot ^{2} ( b )}{cot ( b )} + \frac{1}{cot ^{2} ( b )} = 0

令

: cot (b) = u

1 - \frac{1 + u ^{2}}{u} + \frac{1}{u ^{2}} = 0

1 - \frac{1 + u ^{2}}{u} + \frac{1}{u ^{2}} = 0 : u = 1, u = i, u = - i

1 - \frac{1 + u ^{2}}{u} + \frac{1}{u ^{2}} = 0

乘以最小公倍数

1 - \frac{1 + u ^{2}}{u} + \frac{1}{u ^{2}} = 0

找到

u, u^{2}

的最小公倍数:

u^{2}

u, u^{2}

最小公倍数 (LCM)

计算出由出现在

u

或

u^{2}

中的因子组成的表达式

= u^{2}

乘以最小公倍数=

u^{2}

1 \cdot u^{2} - \frac{1 + u ^{2}}{u} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

化简

1 \cdot u^{2} - \frac{1 + u ^{2}}{u} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

化简

1 \cdot u^{2} : u^{2}

1 \cdot u^{2}

乘以:

1 \cdot u^{2} = u^{2}

= u^{2}

化简

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

- \frac{1 + u ^{2}}{u} u^{2}

分式相乘:

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

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

约分:

u

= - u (u^{2} + 1)

化简

\frac{1}{u ^{2}} u^{2} : 1

\frac{1}{u ^{2}} u^{2}

分式相乘:

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

= \frac{1 \cdot u ^{2}}{u ^{2}}

约分:

u^{2}

= 1

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

u^{2} - u (u^{2} + 1) + 1 = 0

u^{2} - u (u^{2} + 1) + 1 = 0

u^{2} - u (u^{2} + 1) + 1 = 0

解

u^{2} - u (u^{2} + 1) + 1 = 0 : u = 1, u = i, u = - i

u^{2} - u (u^{2} + 1) + 1 = 0

展开

u^{2} - u (u^{2} + 1) + 1 : u^{2} - u^{3} - u + 1

u^{2} - u (u^{2} + 1) + 1

乘开

- u (u^{2} + 1) : - u^{3} - u

- u (u^{2} + 1)

使用分配律:

a (b + c) = ab + a c

a = - u, b = u^{2}, c = 1

= - u u^{2} + (- u) \cdot 1

使用加减运算法则

+ (- a) = - a

= - u^{2} u - 1 \cdot u

化简

- u^{2} u - 1 \cdot u : - u^{3} - u

- u^{2} u - 1 \cdot u

u^{2} u = u^{3}

u^{2} u

使用指数法则:

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

u^{2} u = u^{2 + 1}

= u^{2 + 1}

数字相加:

2 + 1 = 3

= u^{3}

1 \cdot u = u

1 \cdot u

乘以:

1 \cdot u = u

= u

= - u^{3} - u

= - u^{3} - u

= u^{2} - u^{3} - u + 1

u^{2} - u^{3} - u + 1 = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + b = 0

- u^{3} + u^{2} - u + 1 = 0

因式分解

- u^{3} + u^{2} - u + 1 : - (u - 1) (u^{2} + 1)

- u^{3} + u^{2} - u + 1

因式分解出通项

- 1

= - (u^{3} - u^{2} + u - 1)

分解

u^{3} - u^{2} + u - 1 : (u - 1) (u^{2} + 1)

u^{3} - u^{2} + u - 1

= (u^{3} - u^{2}) + (u - 1)

从

u^{3} - u^{2}

分解出因式

u^{2}

:

u^{2} (u - 1)

u^{3} - u^{2}

使用指数法则:

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

u^{3} = u u^{2}

= u u^{2} - u^{2}

因式分解出通项

u^{2}

= u^{2} (u - 1)

= (u - 1) + u^{2} (u - 1)

因式分解出通项

u - 1

= (u - 1) (u^{2} + 1)

= - (u - 1) (u^{2} + 1)

- (u - 1) (u^{2} + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u - 1 = 0 or u^{2} + 1 = 0

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解

u^{2} + 1 = 0 : u = i, u = - i

u^{2} + 1 = 0

将

1

到右边

u^{2} + 1 = 0

两边减去

1

u^{2} + 1 - 1 = 0 - 1

化简

u^{2} = - 1

u^{2} = - 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = - 1, u = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

u = i, u = - i

解为

u = 1, u = i, u = - i

u = 1, u = i, u = - i

验证解

找到无定义的点（奇点）:

u = 0

取

1 - \frac{1 + u ^{2}}{u} + \frac{1}{u ^{2}}

的分母，令其等于零

u = 0

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 1, u = i, u = - i

u = cot (b)

代回

cot (b) = 1, cot (b) = i, cot (b) = - i

cot (b) = 1, cot (b) = i, cot (b) = - i

cot (b) = 1 : b = \frac{π}{4} + πn

cot (b) = 1

cot (b) = 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

b = \frac{π}{4} + πn

b = \frac{π}{4} + πn

cot (b) = i :

无解

cot (b) = i

无解

cot (b) = - i :

无解

cot (b) = - i

无解

合并所有解

b = \frac{π}{4} + πn

作图

流行的例子

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