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人気のある三角関数 >

証明する 1/(sec(x)(sec(x)-cos(x)))=cot^2(x)

解

証明する

\frac{1}{sec ( x ) ( sec ( x ) - cos ( x ) )} = cot^{2} (x)

解

真

解答ステップ

\frac{1}{sec ( x ) ( sec ( x ) - cos ( x ) )} = cot^{2} (x)

左側を操作する

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

サイン, コサインで表わす

\frac{1}{( - cos ( x ) + sec ( x ) ) sec ( x )}

基本的な三角関数の公式を使用する:

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

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

簡素化

\frac{1}{( - cos ( x ) + \frac{1}{c o s ( x )} ) \frac{1}{c o s ( x )}} : \frac{cos ^{2} ( x )}{- cos ^{2} ( x ) + 1}

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

乗じる

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

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

分数を乗じる:

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

= \frac{1 \cdot ( - cos ( x ) + \frac{1}{c o s ( x )} )}{cos ( x )}

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

1 \cdot (- cos (x) + \frac{1}{cos ( x )})

乗算:

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

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

括弧を削除する:

(- a) = - a

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

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

結合

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

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

元を分数に変換する:

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

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

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{- cos ( x ) cos ( x ) + 1}{cos ( x )}

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

- cos (x) cos (x) + 1

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

cos (x) cos (x)

指数の規則を適用する:

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

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

= cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= cos^{2} (x)

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

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

= \frac{\frac{- c o s ^{2} ( x ) + 1}{c o s ( x )}}{cos ( x )}

分数の規則を適用する:

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

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

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

cos (x) cos (x)

指数の規則を適用する:

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

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

= cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= cos^{2} (x)

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

= \frac{1}{\frac{- c o s ^{2} ( x ) + 1}{c o s ^{2} ( x )}}

分数の規則を適用する:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

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

右側を操作する

cot^{2} (x)

サイン, コサインで表わす

cot^{2} (x)

基本的な三角関数の公式を使用する:

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

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

簡素化

(\frac{cos ( x )}{sin ( x )})^{2} : \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

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

指数の規則を適用する:

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

両辺を同じ形式にできることを証明した

\Rightarrow 真

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