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受欢迎的三角函数 >

证明 sec(b)+tan(b)=(cos(b))/(1-sin(b))

解答

证明

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

解答

真

求解步骤

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

调整左侧

sec (b) + tan (b)

用 sin, cos 表示

sec (b) + tan (b)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

使用法则

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

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

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

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

使用三角恒等式改写

乘以

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

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

乘开

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

(1 + sin (b)) (1 - sin (b))

使用平方差公式:

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

a = 1, b = sin (b)

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (b)

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

使用毕达哥拉斯恒等式:

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

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

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

约分:

cos (b)

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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