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受欢迎的三角函数 >

证明 1/(sec^2(θ))+1/(csc^2(θ))=1

解答

证明

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

解答

真

求解步骤

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

调整左侧

\frac{1}{sec ^{2} ( θ )} + \frac{1}{csc ^{2} ( θ )}

用 sin, cos 表示

\frac{1}{csc ^{2} ( θ )} + \frac{1}{sec ^{2} ( θ )}

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{( \frac{1}{s i n ( θ )} ) ^{2}} + \frac{1}{sec ^{2} ( θ )}

使用基本三角恒等式:

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

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

化简

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

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

\frac{1}{( \frac{1}{s i n ( θ )} ) ^{2}} = sin^{2} (θ)

\frac{1}{( \frac{1}{s i n ( θ )} ) ^{2}}

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

(\frac{1}{sin ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{\frac{1}{s i n ^{2} ( θ )}}

使用分式法则:

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

= \frac{sin ^{2} ( θ )}{1}

使用法则

\frac{a}{1} = a

= sin^{2} (θ)

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

\frac{1}{( \frac{1}{c o s ( θ )} ) ^{2}}

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

(\frac{1}{cos ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{\frac{1}{c o s ^{2} ( θ )}}

使用分式法则:

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

= \frac{cos ^{2} ( θ )}{1}

使用法则

\frac{a}{1} = a

= cos^{2} (θ)

= sin^{2} (θ) + cos^{2} (θ)

= sin^{2} (θ) + cos^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

使用三角恒等式改写

cos^{2} (θ) + sin^{2} (θ)

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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