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受欢迎的三角函数 >

证明-tan(t)+(cos(t))/(1-sin(t))=sec(t)

解答

证明

- tan (t) + \frac{cos ( t )}{1 - sin ( t )} = sec (t)

解答

真

求解步骤

- tan (t) + \frac{cos ( t )}{1 - sin ( t )} = sec (t)

调整左侧

- tan (t) + \frac{cos ( t )}{1 - sin ( t )}

用 sin, cos 表示

\frac{cos ( t )}{1 - sin ( t )} - tan (t)

使用基本三角恒等式:

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

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

化简

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

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

1 - sin (t), cos (t)

的最小公倍数:

cos (t) (- sin (t) + 1)

1 - sin (t), cos (t)

最小公倍数 (LCM)

计算出由出现在

1 - sin (t)

或

cos (t)

中的因子组成的表达式

= cos (t) (- sin (t) + 1)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos (t) (- sin (t) + 1)

对于

\frac{cos ( t )}{1 - sin ( t )} :

将分母和分子乘以

cos (t)

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

对于

\frac{sin ( t )}{cos ( t )} :

将分母和分子乘以

- sin (t) + 1

\frac{sin ( t )}{cos ( t )} = \frac{sin ( t ) ( - sin ( t ) + 1 )}{cos ( t ) ( - sin ( t ) + 1 )}

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

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{cos ^{2} ( t ) - sin ( t ) ( - sin ( t ) + 1 )}{cos ( t ) ( - sin ( t ) + 1 )}

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

乘开

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

1 - sin^{2} (t) - (1 - sin (t)) sin (t)

= 1 - sin^{2} (t) - sin (t) (1 - sin (t))

乘开

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

- sin (t) (1 - sin (t))

使用分配律:

a (b - c) = ab - a c

a = - sin (t), b = 1, c = sin (t)

= - sin (t) \cdot 1 - (- sin (t)) sin (t)

使用加减运算法则

- (- a) = a

= - 1 \cdot sin (t) + sin (t) sin (t)

化简

- 1 \cdot sin (t) + sin (t) sin (t) : - sin (t) + sin^{2} (t)

- 1 \cdot sin (t) + sin (t) sin (t)

1 \cdot sin (t) = sin (t)

1 \cdot sin (t)

乘以:

1 \cdot sin (t) = sin (t)

= sin (t)

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

sin (t) sin (t)

使用指数法则:

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

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

= sin^{1 + 1} (t)

数字相加:

1 + 1 = 2

= sin^{2} (t)

= - sin (t) + sin^{2} (t)

= - sin (t) + sin^{2} (t)

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

化简

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

1 - sin^{2} (t) - sin (t) + sin^{2} (t)

对同类项分组

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

同类项相加:

- sin^{2} (t) + sin^{2} (t) = 0

= - sin (t) + 1

= - sin (t) + 1

= \frac{- sin ( t ) + 1}{cos ( t ) ( - sin ( t ) + 1 )}

约分:

- sin (t) + 1

= \frac{1}{cos ( t )}

= \frac{1}{cos ( t )}

= \frac{1}{cos ( t )}

使用三角恒等式改写

使用基本三角恒等式:

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

\frac{1}{\frac{1}{s e c ( t )}}

化简

\frac{1}{\frac{1}{s e c ( t )}}

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= sec (t)

sec (t)

sec (t)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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

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