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受欢迎的三角函数 >

证明 1/(1+tan^2(x))+1/(1+cot^2(x))=1

解答

证明

\frac{1}{1 + tan ^{2} ( x )} + \frac{1}{1 + cot ^{2} ( x )} = 1

解答

真

求解步骤

\frac{1}{1 + tan ^{2} ( x )} + \frac{1}{1 + cot ^{2} ( x )} = 1

调整左侧

\frac{1}{1 + tan ^{2} ( x )} + \frac{1}{1 + cot ^{2} ( x )}

用 sin, cos 表示

\frac{1}{1 + cot ^{2} ( x )} + \frac{1}{1 + tan ^{2} ( x )}

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

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

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

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

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

乘以:

1 \cdot sin^{2} (x) = sin^{2} (x)

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

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

使用分式法则:

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

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

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

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

乘以:

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

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

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

使用分式法则:

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

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

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

使用法则

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

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

使用法则

\frac{a}{a} = 1

= 1

= 1

= 1

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

sin (3 x) = 1

cos(2x)+cos(x)=0

cos (2 x) + cos (x) = 0

csch (ln (2))

tan(2x)=sqrt(3)

tan (2 x) = 3

solvefor x,tan(x)+sec(x)=2cos(x)

so l v e f or x, tan (x) + sec (x) = 2 cos (x)