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受欢迎的三角函数 >

证明 (1+csc(β))/(cot(β)+cos(β))=sec(β)

解答

证明

\frac{1 + csc ( β )}{cot ( β ) + cos ( β )} = sec (β)

解答

真

求解步骤

\frac{1 + csc ( β )}{cot ( β ) + cos ( β )} = sec (β)

调整左侧

\frac{1 + csc ( β )}{cot ( β ) + cos ( β )}

用 sin, cos 表示

\frac{1 + csc ( β )}{cos ( β ) + cot ( β )}

使用基本三角恒等式:

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

= \frac{1 + \frac{1}{s i n ( β )}}{cos ( β ) + cot ( β )}

使用基本三角恒等式:

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

= \frac{1 + \frac{1}{s i n ( β )}}{cos ( β ) + \frac{c o s ( β )}{s i n ( β )}}

化简

\frac{1 + \frac{1}{s i n ( β )}}{cos ( β ) + \frac{c o s ( β )}{s i n ( β )}} : \frac{1}{cos ( β )}

\frac{1 + \frac{1}{s i n ( β )}}{cos ( β ) + \frac{c o s ( β )}{s i n ( β )}}

化简

cos (β) + \frac{cos ( β )}{sin ( β )} : \frac{cos ( β ) sin ( β ) + cos ( β )}{sin ( β )}

cos (β) + \frac{cos ( β )}{sin ( β )}

将项转换为分式:

cos (β) = \frac{cos ( β ) sin ( β )}{sin ( β )}

= \frac{cos ( β ) sin ( β )}{sin ( β )} + \frac{cos ( β )}{sin ( β )}

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{cos ( β ) sin ( β ) + cos ( β )}{sin ( β )}

= \frac{1 + \frac{1}{s i n ( β )}}{\frac{c o s ( β ) s i n ( β ) + c o s ( β )}{s i n ( β )}}

化简

1 + \frac{1}{sin ( β )} : \frac{sin ( β ) + 1}{sin ( β )}

1 + \frac{1}{sin ( β )}

将项转换为分式:

1 = \frac{1 sin ( β )}{sin ( β )}

= \frac{1 \cdot sin ( β )}{sin ( β )} + \frac{1}{sin ( β )}

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 sin ( β ) + 1}{sin ( β )}

乘以:

1 \cdot sin (β) = sin (β)

= \frac{sin ( β ) + 1}{sin ( β )}

= \frac{\frac{s i n ( β ) + 1}{s i n ( β )}}{\frac{c o s ( β ) s i n ( β ) + c o s ( β )}{s i n ( β )}}

分式相除:

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

= \frac{( sin ( β ) + 1 ) sin ( β )}{sin ( β ) ( cos ( β ) sin ( β ) + cos ( β ) )}

约分:

sin (β)

= \frac{sin ( β ) + 1}{cos ( β ) sin ( β ) + cos ( β )}

因式分解出通项

cos (β)

= \frac{sin ( β ) + 1}{cos ( β ) ( sin ( β ) + 1 )}

约分:

sin (β) + 1

= \frac{1}{cos ( β )}

= \frac{1}{cos ( β )}

= \frac{1}{cos ( β )}

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 ( β )}}

化简

\frac{1}{\frac{1}{s e c ( β )}}

使用分式法则:

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

= \frac{sec ( β )}{1}

使用法则

\frac{a}{1} = a

= sec (β)

sec (β)

sec (β)

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

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