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受欢迎的三角函数 >

证明 (1+tan^2(θ))/(1-tan^2(θ))=sec(2θ)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

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

乘以:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

化简

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

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

将项转换为分式:

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

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

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

乘以:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

分式相除:

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

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

约分:

cos^{2} (θ)

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

= \frac{1}{cos ( 2 θ )}

= \frac{1}{cos ( 2 θ )}

使用三角恒等式改写

使用基本三角恒等式:

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

\frac{1}{\frac{1}{s e c ( 2 θ )}}

化简

\frac{1}{\frac{1}{s e c ( 2 θ )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( 2 θ )}{1}

使用法则

\frac{a}{1} = a

= sec (2 θ)

sec (2 θ)

sec (2 θ)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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