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受欢迎的三角函数 >

证明 sec(x)-tan(x)(sin(x))= 1/(sec(x))

解答

证明

sec (x) - tan (x) (sin (x)) = \frac{1}{sec ( x )}

解答

真

求解步骤

sec (x) - tan (x) sin (x) = \frac{1}{sec ( x )}

调整左侧

sec (x) - tan (x) sin (x)

用 sin, cos 表示

sec (x) - sin (x) tan (x)

使用基本三角恒等式:

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

= \frac{1}{cos ( x )} - sin (x) tan (x)

使用基本三角恒等式:

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

= \frac{1}{cos ( x )} - sin (x) \frac{sin ( x )}{cos ( x )}

化简

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

\frac{1}{cos ( x )} - sin (x) \frac{sin ( x )}{cos ( x )}

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

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

分式相乘:

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

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

使用指数法则:

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

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

使用法则

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

1 - sin^{2} (x) = cos^{2} (x)

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

约分:

cos (x)

= cos (x)

= cos (x)

使用基本三角恒等式:

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

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

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 (1-tan^4(x))/(sec^2(x))=1-tan^2(x)

p ro v e \frac{1 - tan ^{4} ( x )}{sec ^{2} ( x )} = 1 - tan^{2} (x)

证明 3-sec^2(x)=2-tan^2(x)

p ro v e 3 - sec^{2} (x) = 2 - tan^{2} (x)

证明 (cot(x)+tan(x))^2=csc^2(x)sec^2(x)

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

证明 cos(3x)=4cos(3x)-3cos(x)

p ro v e cos (3 x) = 4 cos (3 x) - 3 cos (x)

证明 sec^2(a)+csc^2(a)=sec^2(a)csc^2(a)

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