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人気のある三角関数 >

証明する (cot(x)+tan(x))^2=sec^2(x)csc^2(x)

解

証明する

(cot (x) + tan (x))^{2} = sec^{2} (x) csc^{2} (x)

解

真

解答ステップ

(cot (x) + tan (x))^{2} = sec^{2} (x) csc^{2} (x)

左側を操作する

(cot (x) + tan (x))^{2}

サイン, コサインで表わす

(cot (x) + tan (x))^{2}

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 )} + tan (x))^{2}

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

結合

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

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

以下の最小公倍数:

sin (x), cos (x) : sin (x) cos (x)

sin (x), cos (x)

最小公倍数 (LCM)

sin (x)

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:

cos (x)

= sin (x) cos (x)

LCMに基づいて分数を調整する

該当する分母を乗じてLCMに変えるために

必要な量で各分子を乗じる

sin (x) cos (x)

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

の場合

:

分母と分子に以下を乗じる:

cos (x)

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

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

の場合

:

分母と分子に以下を乗じる:

sin (x)

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

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

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

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

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

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

三角関数の公式を使用して書き換える

基本的な三角関数の公式を使用する:

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

\frac{1}{cos ^{2} ( x ) ( \frac{1}{c s c ( x )} ) ^{2}}

基本的な三角関数の公式を使用する:

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

\frac{1}{( \frac{1}{s e c ( x )} ) ^{2} ( \frac{1}{c s c ( x )} ) ^{2}}

簡素化

\frac{1}{( \frac{1}{s e c ( x )} ) ^{2} ( \frac{1}{c s c ( x )} ) ^{2}}

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

(\frac{1}{sec ( x )})^{2}

指数の規則を適用する:

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

= \frac{1}{( \frac{1}{c s c ( x )} ) ^{2} \frac{1}{s e c ^{2} ( x )}}

(\frac{1}{csc ( x )})^{2} = \frac{1}{csc ^{2} ( x )}

(\frac{1}{csc ( x )})^{2}

指数の規則を適用する:

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

= \frac{1 ^{2}}{csc ^{2} ( x )}

規則を適用

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( x )}

= \frac{1}{\frac{1}{s e c ^{2} ( x )} \cdot \frac{1}{c s c ^{2} ( x )}}

乗じる

\frac{1}{sec ^{2} ( x )} \cdot \frac{1}{csc ^{2} ( x )} : \frac{1}{sec ^{2} ( x ) csc ^{2} ( x )}

\frac{1}{sec ^{2} ( x )} \cdot \frac{1}{csc ^{2} ( x )}

分数を乗じる:

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

= \frac{1 \cdot 1}{sec ^{2} ( x ) csc ^{2} ( x )}

数を乗じる:

1 \cdot 1 = 1

= \frac{1}{sec ^{2} ( x ) csc ^{2} ( x )}

= \frac{1}{\frac{1}{s e c ^{2} ( x ) c s c ^{2} ( x )}}

分数の規則を適用する:

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

= \frac{sec ^{2} ( x ) csc ^{2} ( x )}{1}

規則を適用

\frac{a}{1} = a

= sec^{2} (x) csc^{2} (x)

sec^{2} (x) csc^{2} (x)

sec^{2} (x) csc^{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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