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受欢迎的三角函数 >

证明 1/(1-sin(B))=sec^2(B)+sec(B)tan(B)

解答

证明

\frac{1}{1 - sin ( B )} = sec^{2} (B) + sec (B) tan (B)

解答

真

求解步骤

\frac{1}{1 - sin ( B )} = sec^{2} (B) + sec (B) tan (B)

调整右侧

sec^{2} (B) + sec (B) tan (B)

用 sin, cos 表示

sec^{2} (B) + sec (B) tan (B)

使用基本三角恒等式:

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

= (\frac{1}{cos ( B )})^{2} + \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 )})^{2} + \frac{1}{cos ( B )} \cdot \frac{sin ( B )}{cos ( B )}

化简

(\frac{1}{cos ( B )})^{2} + \frac{1}{cos ( B )} \cdot \frac{sin ( B )}{cos ( B )} : \frac{1 + sin ( B )}{cos ^{2} ( B )}

(\frac{1}{cos ( B )})^{2} + \frac{1}{cos ( B )} \cdot \frac{sin ( B )}{cos ( B )}

(\frac{1}{cos ( B )})^{2} = \frac{1}{cos ^{2} ( B )}

(\frac{1}{cos ( B )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{cos ^{2} ( B )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( B )}

\frac{1}{cos ( B )} \cdot \frac{sin ( B )}{cos ( B )} = \frac{sin ( B )}{cos ^{2} ( B )}

\frac{1}{cos ( B )} \cdot \frac{sin ( B )}{cos ( B )}

分式相乘:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot sin ( B )}{cos ( B ) cos ( B )}

乘以:

1 \cdot sin (B) = sin (B)

= \frac{sin ( B )}{cos ( B ) cos ( B )}

cos (B) cos (B) = cos^{2} (B)

cos (B) cos (B)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (B) cos (B) = cos^{1 + 1} (B)

= cos^{1 + 1} (B)

数字相加:

1 + 1 = 2

= cos^{2} (B)

= \frac{sin ( B )}{cos ^{2} ( B )}

= \frac{1}{cos ^{2} ( B )} + \frac{sin ( B )}{cos ^{2} ( B )}

使用法则

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

= \frac{1 + sin ( B )}{cos ^{2} ( B )}

= \frac{1 + sin ( B )}{cos ^{2} ( B )}

= \frac{1 + sin ( B )}{cos ^{2} ( B )}

使用三角恒等式改写

\frac{1 + sin ( B )}{cos ^{2} ( B )}

使用毕达哥拉斯恒等式:

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

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

= \frac{1 + sin ( B )}{1 - sin ^{2} ( B )}

分解

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

1 - sin^{2} (B)

使用平方差公式:

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

1 - sin^{2} (B) = (1 + sin (B)) (1 - sin (B))

= (1 + sin (B)) (1 - sin (B))

= \frac{1 + sin ( B )}{( 1 + sin ( B ) ) ( 1 - sin ( B ) )}

约分:

1 + sin (B)

= \frac{1}{1 - sin ( B )}

= \frac{1}{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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