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受欢迎的三角函数 >

证明 tan^2(t)= 1/(cos^2(t))-1

解答

证明

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

解答

真

求解步骤

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

调整左侧

tan^{2} (t)

用 sin, cos 表示

tan^{2} (t)

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

乘开

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

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

使用分式法则:

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

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

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

使用法则

\frac{a}{a} = 1

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

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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