ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

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

解

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

解

b = \frac{π}{4} + πn

+1

度

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

解答ステップ

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

両辺から

csc^{2} (b)

を引く

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

簡素化

\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)

元を分数に変換する:

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 )}

分母が等しいので, 分数を組み合わせる:

\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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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}

指数の規則を適用する:

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

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

規則を適用

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 )}

分数を乗じる:

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))

乗算:

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

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

括弧を削除する:

(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

置換で解く

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

仮定:

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

LCMで乗じる

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

以下の最小公倍数を求める:

u, u^{2} : u^{2}

u, u^{2}

最小公倍数 (LCM)

u

または以下のいずれかに現れる因数で構成された式を計算する:

u^{2}

= u^{2}

以下で乗じる: LCM=

u^{2}

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

簡素化

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

簡素化

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

1 \cdot u^{2}

乗算:

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

= u^{2}

簡素化

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

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

分数を乗じる:

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

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

共通因数を約分する:

u

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

簡素化

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

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

分数を乗じる:

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

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

共通因数を約分する:

u^{2}

= 1

簡素化

0 \cdot u^{2} : 0

0 \cdot u^{2}

規則を適用

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

解く

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

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

拡張

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

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

拡張

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

- u (u^{2} + 1)

分配法則を適用する:

a (b + c) = ab + a c

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

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

マイナス・プラスの規則を適用する

+ (- a) = - a

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

簡素化

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

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

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

u^{2} u

指数の規則を適用する:

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

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

= u^{2 + 1}

数を足す:

2 + 1 = 3

= u^{3}

1 \cdot u = u

1 \cdot u

乗算:

1 \cdot u = u

= u

= - u^{3} - u

= - u^{3} - u

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

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

標準的な形式で書く

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

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

因数

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

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

共通項をくくり出す

- 1

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

因数

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

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

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

u^{2}

を

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

からくくり出す

u^{3} - u^{2}

指数の規則を適用する:

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

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

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

共通項をくくり出す

u^{2}

= u^{2} (u - 1)

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

共通項をくくり出す

u - 1

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

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

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

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

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

解く

u - 1 = 0 : u = 1

u - 1 = 0

1

を右側に移動します

u - 1 = 0

両辺に

1

を足す

u - 1 + 1 = 0 + 1

簡素化

u = 1

u = 1

解く

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

u^{2} + 1 = 0

1

を右側に移動します

u^{2} + 1 = 0

両辺から

1

を引く

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

簡素化

u^{2} = - 1

u^{2} = - 1

x^{2} = f (a)

の場合, 解は

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

u = - 1, u = - - 1

簡素化

- 1 : i

- 1

虚数の規則を適用する:

- 1 = i

= i

簡素化

- - 1 : - i

- - 1

虚数の規則を適用する:

- 1 = i

= - i

u = i, u = - i

解答は

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

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

解を検算する

未定義の (特異) 点を求める:

u = 0

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

の分母をゼロに比較する

u = 0

解く

u^{2} = 0 : u = 0

u^{2} = 0

規則を適用

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

u = 0

以下の点は定義されていない

u = 0

未定義のポイントを解に組み合わせる:

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

代用を戻す

u = cot (b)

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

以下の一般解

cot (b) = 1

cot (x)

πn

循環を含む周期性テーブル:

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 :

解なし

cot (b) = i

解なし

cot (b) = - i :

解なし

cot (b) = - i

解なし

すべての解を組み合わせる

b = \frac{π}{4} + πn

グラフ

人気の例

solvefor x,r+s+6t=cos(2x+y)

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

sin(x)+sin^2(x/2)= 1/2

sin^5(x)+sin^3(x)=0

5sin^2(x)cos(7x)-cos(7x)=0