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受欢迎的三角函数 >

证明 (cos(x))/(1-sin(x))=sec(x)+tan(x)

解答

证明

\frac{cos ( x )}{1 - sin ( x )} = sec (x) + tan (x)

解答

真

求解步骤

\frac{cos ( x )}{1 - sin ( x )} = sec (x) + tan (x)

调整右侧

sec (x) + tan (x)

用 sin, cos 表示

sec (x) + tan (x)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= \frac{1}{cos ( x )} + tan (x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{1}{cos ( x )} + \frac{sin ( x )}{cos ( x )}

化简

\frac{1}{cos ( x )} + \frac{sin ( x )}{cos ( x )} : \frac{1 + sin ( x )}{cos ( x )}

\frac{1}{cos ( x )} + \frac{sin ( x )}{cos ( x )}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + sin ( x )}{cos ( x )}

= \frac{1 + sin ( x )}{cos ( x )}

= \frac{1 + sin ( x )}{cos ( x )}

使用三角恒等式改写

乘以

\frac{1 - sin ( x )}{1 - sin ( x )}

= \frac{( 1 + sin ( x ) ) ( 1 - sin ( x ) )}{( 1 - sin ( x ) ) cos ( x )}

乘开

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

(1 + sin (x)) (1 - sin (x))

使用平方差公式:

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

a = 1, b = sin (x)

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

= \frac{1 - sin ^{2} ( x )}{( 1 - sin ( x ) ) cos ( x )}

使用毕达哥拉斯恒等式:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= \frac{cos ^{2} ( x )}{( 1 - sin ( x ) ) cos ( x )}

约分:

cos (x)

= \frac{cos ( x )}{1 - sin ( x )}

= \frac{cos ( x )}{1 - sin ( x )}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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