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人気のある三角関数 >

証明する (1-tan^4(x))/(1-tan^2(x))=sec^2(x)

解

証明する

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} = sec^{2} (x)

解

真

解答ステップ

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} = sec^{2} (x)

左側を操作する

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )}

簡素化

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} : tan^{2} (x) + 1

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )}

因数

1 - tan^{4} (x) : - (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

1 - tan^{4} (x)

共通項をくくり出す

- 1

= - (tan^{4} (x) - 1)

因数

tan^{4} (x) - 1 : (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

tan^{4} (x) - 1

tan^{4} (x) - 1

を書き換え

(tan^{2} (x))^{2} - 1^{2}

tan^{4} (x) - 1

1

を書き換え

1^{2}

= tan^{4} (x) - 1^{2}

指数の規則を適用する:

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

tan^{4} (x) = (tan^{2} (x))^{2}

= (tan^{2} (x))^{2} - 1^{2}

= (tan^{2} (x))^{2} - 1^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

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

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

因数

tan^{2} (x) - 1 : (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) - 1

1

を書き換え

1^{2}

= tan^{2} (x) - 1^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

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

= (tan (x) + 1) (tan (x) - 1)

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

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

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

因数

1 - tan^{2} (x) : - (tan (x) + 1) (tan (x) - 1)

1 - tan^{2} (x)

共通項をくくり出す

- 1

= - (tan^{2} (x) - 1)

因数

tan^{2} (x) - 1 : (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) - 1

1

を書き換え

1^{2}

= tan^{2} (x) - 1^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

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

= (tan (x) + 1) (tan (x) - 1)

= - (tan (x) + 1) (tan (x) - 1)

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

キャンセル

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

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

共通因数を約分する:

tan (x) + 1

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

共通因数を約分する:

tan (x) - 1

= tan^{2} (x) + 1

= tan^{2} (x) + 1

= 1 + tan^{2} (x)

三角関数の公式を使用して書き換える

1 + tan^{2} (x)

ピタゴラスの公式を使用する:

tan^{2} (x) + 1 = sec^{2} (x)

= sec^{2} (x)

= sec^{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 cos(4x)=1-2sin^2(2x)

p ro v e cos (4 x) = 1 - 2 sin^{2} (2 x)

証明する arcsin(x)=csc(x)

p ro v e arcsin (x) = csc (x)

証明する cos^4(A)-sin^4(A)=cos(2A)

p ro v e cos^{4} (A) - sin^{4} (A) = cos (2 A)

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

p ro v e \frac{1 + sec ^{2} ( x )}{sec ^{2} ( x )} = 1 + cos^{2} (x)

証明する tan(θ)+1=sec(θ)

p ro v e tan (θ) + 1 = sec (θ)