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受欢迎的三角函数 >

证明 1+(tan^2(A))/(1+sec(A))=sec(A)

解答

证明

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )} = sec (A)

解答

真

求解步骤

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )} = sec (A)

调整左侧

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )}

使用三角恒等式改写

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )}

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

消掉

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

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

分解

sec^{2} (A) - 1 : (sec (A) + 1) (sec (A) - 1)

sec^{2} (A) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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

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

= (sec (A) + 1) (sec (A) - 1)

= \frac{( sec ( A ) + 1 ) ( sec ( A ) - 1 )}{1 + sec ( A )}

约分:

sec (A) + 1

= sec (A) - 1

= 1 + sec (A) - 1

对同类项分组

= sec (A) + 1 - 1

1 - 1 = 0

= sec (A)

= sec (A)

= sec (A)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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