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受欢迎的三角函数 >

证明 (csc^4(x)-1)/(1+csc^2(x))=cot^2(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

化简

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

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

分解

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

- 1 + csc^{4} (x)

将

csc^{4} (x) - 1

改写为

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

csc^{4} (x) - 1

将

1

改写为

1^{2}

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

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

分解

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

csc^{2} (x) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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

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

= (csc (x) + 1) (csc (x) - 1)

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

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

约分:

csc^{2} (x) + 1

= (csc (x) + 1) (csc (x) - 1)

= (- 1 + csc (x)) (1 + csc (x))

使用三角恒等式改写

(- 1 + csc (x)) (1 + csc (x))

乘开

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

(csc (x) + 1) (csc (x) - 1)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = csc (x), b = 1

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

使用法则

1^{a} = 1

1^{2} = 1

= csc^{2} (x) - 1

= csc^{2} (x) - 1

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

csc^{2} (x) - 1 = cot^{2} (x)

= cot^{2} (x)

= cot^{2} (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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