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人気のある三角関数 >

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

解

証明する

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

解

真

解答ステップ

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

左側を操作する

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

サイン, コサインで表わす

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

簡素化

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

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

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

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

指数の規則を適用する:

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

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

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

指数の規則を適用する:

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

分数の規則を適用する:

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

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

結合

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

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

元を分数に変換する:

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

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

乗算:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

分数の規則を適用する:

\frac{a}{1} = a

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

分数を乗じる:

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

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

共通因数を約分する:

cos^{2} (x)

= cos^{2} (x) + 1

= cos^{2} (x) + 1

= 1 + cos^{2} (x)

両辺を同じ形式にできることを証明した

\Rightarrow 真

人気の例

証明する tan(θ)+1=sec(θ)

p ro v e tan (θ) + 1 = sec (θ)

証明する cos(θ)sec(θ)-cos^2(θ)=sin^2(θ)

p ro v e cos (θ) sec (θ) - cos^{2} (θ) = sin^{2} (θ)

証明する sin(12x)=2sin(6x)cos(6x)

p ro v e sin (12 x) = 2 sin (6 x) cos (6 x)

証明する 1/(cos(θ))-cos(θ)=sin(θ)tan(θ)

p ro v e \frac{1}{cos ( θ )} - cos (θ) = sin (θ) tan (θ)

証明する 1+tan^2(30)=sec^2(30)

p ro v e 1 + tan^{2} (3 0^{\circ}) = sec^{2} (3 0^{\circ})