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受欢迎的三角函数 >

证明 tan(β)=(cos(β))/(sin(β)cot^2(β))

解答

证明

tan (β) = \frac{cos ( β )}{sin ( β ) cot ^{2} ( β )}

解答

真

求解步骤

tan (β) = \frac{cos ( β )}{sin ( β ) cot ^{2} ( β )}

调整右侧

\frac{cos ( β )}{sin ( β ) cot ^{2} ( β )}

用 sin, cos 表示

\frac{cos ( β )}{cot ^{2} ( β ) sin ( β )}

使用基本三角恒等式:

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

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

化简

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

\frac{cos ( β )}{( \frac{c o s ( β )}{s i n ( β )} ) ^{2} sin ( β )}

(\frac{cos ( β )}{sin ( β )})^{2} = \frac{cos ^{2} ( β )}{sin ^{2} ( β )}

(\frac{cos ( β )}{sin ( β )})^{2}

使用指数法则:

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

= \frac{cos ^{2} ( β )}{sin ^{2} ( β )}

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

乘

\frac{cos ^{2} ( β )}{sin ^{2} ( β )} sin (β) : \frac{cos ^{2} ( β )}{sin ( β )}

\frac{cos ^{2} ( β )}{sin ^{2} ( β )} sin (β)

分式相乘:

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

= \frac{cos ^{2} ( β ) sin ( β )}{sin ^{2} ( β )}

约分:

sin (β)

= \frac{cos ^{2} ( β )}{sin ( β )}

= \frac{cos ( β )}{\frac{c o s ^{2} ( β )}{s i n ( β )}}

使用分式法则:

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

= \frac{cos ( β ) sin ( β )}{cos ^{2} ( β )}

约分:

cos (β)

= \frac{sin ( β )}{cos ( β )}

= \frac{sin ( β )}{cos ( β )}

= \frac{sin ( β )}{cos ( β )}

使用基本三角恒等式:

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

= tan (β)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 (sin^2(x))/(cos(x))=sin(x)tan(x)

p ro v e \frac{sin ^{2} ( x )}{cos ( x )} = sin (x) tan (x)

证明 csc^2(x)sec^2(x)=sec^2(x)+csc^2(x)

p ro v e csc^{2} (x) sec^{2} (x) = sec^{2} (x) + csc^{2} (x)

证明 (sin(x)+cos(x))/(cos(x))=1+tan(x)

p ro v e \frac{sin ( x ) + cos ( x )}{cos ( x )} = 1 + tan (x)

证明 (1+sin(-x))/(cos(x)tan(x)-1)=-1

p ro v e \frac{1 + sin ( - x )}{cos ( x ) tan ( x ) - 1} = - 1

证明 sec(A)-(cos(A))/(1+sin(A))=tan(A)

p ro v e sec (A) - \frac{cos ( A )}{1 + sin ( A )} = tan (A)