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受欢迎的三角函数 >

证明 (1-sin^2(t))(1+tan^2(t))=1

解答

证明

(1 - sin^{2} (t)) (1 + tan^{2} (t)) = 1

解答

真

求解步骤

(1 - sin^{2} (t)) (1 + tan^{2} (t)) = 1

调整左侧

(1 - sin^{2} (t)) (1 + tan^{2} (t))

使用三角恒等式改写

(1 - sin^{2} (t)) (1 + tan^{2} (t))

使用毕达哥拉斯恒等式:

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

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

= cos^{2} (t) (1 + tan^{2} (t))

= cos^{2} (t) (1 + tan^{2} (t))

用 sin, cos 表示

(1 + tan^{2} (t)) cos^{2} (t)

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

因为分母相等，所以合并分式:

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

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

乘以:

1 \cdot cos^{2} (t) = cos^{2} (t)

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

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

分式相乘:

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

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

约分:

cos^{2} (t)

= cos^{2} (t) + sin^{2} (t)

= cos^{2} (t) + sin^{2} (t)

= cos^{2} (t) + sin^{2} (t)

使用三角恒等式改写

cos^{2} (t) + sin^{2} (t)

使用毕达哥拉斯恒等式:

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

= 1

= 1

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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