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受欢迎的三角函数 >

证明 1/(tan(b))+tan(b)=(sec^2(b))/(tan(b))

解答

证明

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

解答

真

求解步骤

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

调整左侧

\frac{1}{tan ( b )} + tan (b)

化简

\frac{1}{tan ( b )} + tan (b) : \frac{1 + tan ^{2} ( b )}{tan ( b )}

\frac{1}{tan ( b )} + tan (b)

将项转换为分式:

tan (b) = \frac{tan ( b ) tan ( b )}{tan ( b )}

= \frac{1}{tan ( b )} + \frac{tan ( b ) tan ( b )}{tan ( b )}

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 + tan ( b ) tan ( b )}{tan ( b )}

1 + tan (b) tan (b) = 1 + tan^{2} (b)

1 + tan (b) tan (b)

tan (b) tan (b) = tan^{2} (b)

tan (b) tan (b)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

tan (b) tan (b) = tan^{1 + 1} (b)

= tan^{1 + 1} (b)

数字相加:

1 + 1 = 2

= tan^{2} (b)

= 1 + tan^{2} (b)

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

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

使用三角恒等式改写

\frac{1 + tan ^{2} ( b )}{tan ( b )}

使用毕达哥拉斯恒等式:

tan^{2} (x) + 1 = sec^{2} (x)

= \frac{sec ^{2} ( b )}{tan ( b )}

= \frac{sec ^{2} ( b )}{tan ( b )}

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 cos^3(x)=(cos(x))^3

p ro v e cos^{3} (x) = (cos (x))^{3}

证明 cos(b)(1+tan(b))^2=sec(b)+2sin(b)

p ro v e cos (b) (1 + tan (b))^{2} = sec (b) + 2 sin (b)

证明 csc^2(x)-csc(x)+9=11

p ro v e csc^{2} (x) - csc (x) + 9 = 11

证明 sin(x/2)=sqrt((1+sin(x))/2)

p ro v e sin (\frac{x}{2}) = \frac{1 + sin ( x )}{2}

证明 (sin^2(x)-1)=-cos^2(x)

p ro v e (sin^{2} (x) - 1) = - cos^{2} (x)