zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 1/(1-sin(r))=sec^2(r)+sec(r)tan(r)

解答

证明

\frac{1}{1 - sin ( r )} = sec^{2} (r) + sec (r) tan (r)

解答

真

求解步骤

\frac{1}{1 - sin ( r )} = sec^{2} (r) + sec (r) tan (r)

调整右侧

sec^{2} (r) + sec (r) tan (r)

用 sin, cos 表示

sec^{2} (r) + sec (r) tan (r)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

(\frac{1}{cos ( r )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

\frac{1}{cos ( r )} \cdot \frac{sin ( r )}{cos ( r )}

分式相乘:

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

= \frac{1 \cdot sin ( r )}{cos ( r ) cos ( r )}

乘以:

1 \cdot sin (r) = sin (r)

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

cos (r) cos (r) = cos^{2} (r)

cos (r) cos (r)

使用指数法则:

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

cos (r) cos (r) = cos^{1 + 1} (r)

= cos^{1 + 1} (r)

数字相加:

1 + 1 = 2

= cos^{2} (r)

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

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

使用法则

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= \frac{1 + sin ( r )}{1 - sin ^{2} ( r )}

分解

1 - sin^{2} (r) : (1 + sin (r)) (1 - sin (r))

1 - sin^{2} (r)

使用平方差公式:

r^{2} - y^{2} = (r + y) (r - y)

1 - sin^{2} (r) = (1 + sin (r)) (1 - sin (r))

= (1 + sin (r)) (1 - sin (r))

= \frac{1 + sin ( r )}{( 1 + sin ( r ) ) ( 1 - sin ( r ) )}

约分:

1 + sin (r)

= \frac{1}{1 - sin ( r )}

= \frac{1}{1 - sin ( r )}

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/(sec(x))=sec(x)-tan(x)sin(x)

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

证明 sin(a+b)sin(a-b)=sin^2(b)-sin^2(a)

p ro v e sin (a + b) sin (a - b) = sin^{2} (b) - sin^{2} (a)

证明 sec(b)+tan(b)=(cos(b))/(1-sin(b))

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

证明 cot^4(x)+cot^2(x)=csc^4(x)-csc^2(x)

p ro v e cot^{4} (x) + cot^{2} (x) = csc^{4} (x) - csc^{2} (x)

证明 1-tanh^2(x)=sech^2(x)

p ro v e 1 - tanh^{2} (x) = sech^{2} (x)