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受欢迎的三角函数 >

证明 (1+tan^2(u))/(1-tan^2(u))= 1/(cos^2(u)-sin^2(u))

解答

证明

\frac{1 + tan ^{2} ( u )}{1 - tan ^{2} ( u )} = \frac{1}{cos ^{2} ( u ) - sin ^{2} ( u )}

解答

真

求解步骤

\frac{1 + tan ^{2} ( u )}{1 - tan ^{2} ( u )} = \frac{1}{cos ^{2} ( u ) - sin ^{2} ( u )}

调整左侧

\frac{1 + tan ^{2} ( u )}{1 - tan ^{2} ( u )}

用 sin, cos 表示

\frac{1 + tan ^{2} ( u )}{1 - tan ^{2} ( u )}

使用基本三角恒等式:

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

= \frac{1 + ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}}{1 - ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}}

化简

\frac{1 + ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}}{1 - ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}} : \frac{cos ^{2} ( u ) + sin ^{2} ( u )}{cos ^{2} ( u ) - sin ^{2} ( u )}

\frac{1 + ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}}{1 - ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}}

使用指数法则:

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

= \frac{1 + ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}}{1 - \frac{s i n ^{2} ( u )}{c o s ^{2} ( u )}}

使用指数法则:

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

= \frac{1 + \frac{s i n ^{2} ( u )}{c o s ^{2} ( u )}}{1 - \frac{s i n ^{2} ( u )}{c o s ^{2} ( u )}}

化简

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

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

将项转换为分式:

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

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

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} ( u ) - sin ^{2} ( u )}{cos ^{2} ( u )}

乘以:

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

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

= \frac{1 + \frac{s i n ^{2} ( u )}{c o s ^{2} ( u )}}{\frac{c o s ^{2} ( u ) - s i n ^{2} ( u )}{c o s ^{2} ( u )}}

化简

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

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

将项转换为分式:

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

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

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

乘以:

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

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

= \frac{\frac{c o s ^{2} ( u ) + s i n ^{2} ( u )}{c o s ^{2} ( u )}}{\frac{c o s ^{2} ( u ) - s i n ^{2} ( u )}{c o s ^{2} ( u )}}

分式相除:

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

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

约分:

cos^{2} (u)

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

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

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

使用三角恒等式改写

\frac{cos ^{2} ( u ) + sin ^{2} ( u )}{cos ^{2} ( u ) - sin ^{2} ( u )}

使用毕达哥拉斯恒等式:

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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p ro v e 1 + tan^{2} (- x) = sec^{2} (x)

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p ro v e tan (\frac{π}{2} - x) = cot (x)

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p ro v e cos (\frac{π}{4} - x) + cos (\frac{π}{4} + x) = 2 cos (x)

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p ro v e \frac{cos ( x )}{tan ( x )} = csc (x) - sin (x)