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人気のある三角関数 >

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

解

証明する

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

解

真

解答ステップ

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

左側を操作する

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

簡素化

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

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

因数

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

1 - 4 sec^{2} (x)

共通項をくくり出す

- 1

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

因数

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

4 sec^{2} (x) - 1

4 sec^{2} (x) - 1

を書き換え

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

4 sec^{2} (x) - 1

4

を書き換え

2^{2}

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

1

を書き換え

1^{2}

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

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

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

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

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

2乗の差の公式を適用する:

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

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

= (2 sec (x) + 1) (2 sec (x) - 1)

= - (2 sec (x) + 1) (2 sec (x) - 1)

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

共通因数を約分する:

2 sec (x) + 1

= - (2 sec (x) - 1)

否定

- (2 sec (x) - 1) = - 2 sec (x) + 1

= - 2 sec (x) + 1

= 1 - 2 sec (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(x)=cot(pi/2-x)

p ro v e tan (x) = cot (\frac{π}{2} - x)

証明する sec(2x)=(csc^2(x))/(csc^2(x)-2)

p ro v e sec (2 x) = \frac{csc ^{2} ( x )}{csc ^{2} ( x ) - 2}

証明する (1-cos(a))/(1+cos(a))=tan^2(a/2)

p ro v e \frac{1 - cos ( a )}{1 + cos ( a )} = tan^{2} (\frac{a}{2})

証明する csc^2(x)-cos(x)sec(x)=cot^2(x)

p ro v e csc^{2} (x) - cos (x) sec (x) = cot^{2} (x)

証明する (csc(x)+1)/(cot(x)+cos(x))=sec(x)

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