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受欢迎的三角函数 >

证明 (cos(t))/(1-sin(t))=sec(t)+tan(t)

解答

证明

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

解答

真

求解步骤

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

调整右侧

sec (t) + tan (t)

用 sin, cos 表示

sec (t) + tan (t)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

使用法则

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

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

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

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

使用三角恒等式改写

乘以

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

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

乘开

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

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

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sin (t)

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (t)

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

使用毕达哥拉斯恒等式:

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

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

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

约分:

cos (t)

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

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

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(2x)=2(cos(x))/(csc(x))

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

证明 2sin(30)cos(30)=sin(60)

p ro v e 2 sin (3 0^{\circ}) cos (3 0^{\circ}) = sin (6 0^{\circ})

证明 sin(2x)=(2sin(x))/(sec(x))

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

证明 tan(2θ)-tan(θ)=tan(θ)sec(2θ)

p ro v e tan (2 θ) - tan (θ) = tan (θ) sec (2 θ)

证明 tan(2A)-tan(A)=tan(A)sec(2A)

p ro v e tan (2 A) - tan (A) = tan (A) sec (2 A)