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受欢迎的三角函数 >

证明 1+tan^2(α)= 1/(cos^2(α))

解答

证明

1 + tan^{2} (α) = \frac{1}{cos ^{2} ( α )}

解答

真

求解步骤

1 + tan^{2} (α) = \frac{1}{cos ^{2} ( α )}

调整左侧

1 + tan^{2} (α)

用 sin, cos 表示

1 + tan^{2} (α)

使用基本三角恒等式:

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

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

化简

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

1 + (\frac{sin ( α )}{cos ( α )})^{2}

使用指数法则:

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

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

将项转换为分式:

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

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

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

乘以:

1 \cdot cos^{2} (α) = cos^{2} (α)

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

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

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

使用三角恒等式改写

\frac{cos ^{2} ( α ) + sin ^{2} ( α )}{cos ^{2} ( α )}

使用毕达哥拉斯恒等式:

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

= \frac{1}{cos ^{2} ( α )}

= \frac{1}{cos ^{2} ( α )}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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