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受欢迎的三角函数 >

证明 (1-tan^4(x))/(1-tan^2(x))=sec^2(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

化简

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

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

分解

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

1 - tan^{4} (x)

因式分解出通项

- 1

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

分解

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

tan^{4} (x) - 1

将

tan^{4} (x) - 1

改写为

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

tan^{4} (x) - 1

将

1

改写为

1^{2}

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

使用指数法则:

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}

使用平方差公式:

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)

分解

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

tan^{2} (x) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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 )}

分解

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

1 - tan^{2} (x)

因式分解出通项

- 1

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

分解

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

tan^{2} (x) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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 )}

消掉

\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 )}

约分:

tan (x) + 1

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

约分:

tan (x) - 1

= tan^{2} (x) + 1

= tan^{2} (x) + 1

= 1 + tan^{2} (x)

使用三角恒等式改写

1 + tan^{2} (x)

使用毕达哥拉斯恒等式:

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

= sec^{2} (x)

= sec^{2} (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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

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证明 tan(θ)+1=sec(θ)

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