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人気のある三角関数 >

証明する ((sec(θ))/(cos(θ)))-((tan(θ))/(cot(θ)))=1

解

証明する

(\frac{sec ( θ )}{cos ( θ )}) - (\frac{tan ( θ )}{cot ( θ )}) = 1

解

真

解答ステップ

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )} = 1

左側を操作する

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

サイン, コサインで表わす

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

基本的な三角関数の公式を使用する:

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

= \frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

基本的な三角関数の公式を使用する:

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

= \frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{cot ( θ )}

基本的な三角関数の公式を使用する:

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

= \frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

簡素化

\frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}} : \frac{1 - sin ^{2} ( θ )}{cos ^{2} ( θ )}

\frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

\frac{\frac{1}{c o s ( θ )}}{cos ( θ )} = \frac{1}{cos ^{2} ( θ )}

\frac{\frac{1}{c o s ( θ )}}{cos ( θ )}

分数の規則を適用する:

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

= \frac{1}{cos ( θ ) cos ( θ )}

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

指数の規則を適用する:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数を足す:

1 + 1 = 2

= cos^{2} (θ)

= \frac{1}{cos ^{2} ( θ )}

\frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}} = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

\frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

分数を割る:

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

= \frac{sin ( θ ) sin ( θ )}{cos ( θ ) cos ( θ )}

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

指数の規則を適用する:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

数を足す:

1 + 1 = 2

= sin^{2} (θ)

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

指数の規則を適用する:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数を足す:

1 + 1 = 2

= cos^{2} (θ)

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

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

規則を適用

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

= \frac{cos ^{2} ( θ )}{cos ^{2} ( θ )}

規則を適用

\frac{a}{a} = 1

= 1

= 1

両辺を同じ形式にできることを証明した

\Rightarrow 真

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