证明 sec(x)+tan(x)= 1/((sec(x)-tan(x)))

解答

证明

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

真

求解步骤

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

调整右侧

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= \frac{sec ^{2} ( x ) - tan ^{2} ( x )}{sec ( x ) - tan ( x )}

化简

\frac{sec ^{2} ( x ) - tan ^{2} ( x )}{sec ( x ) - tan ( x )} : sec (x) + tan (x)

\frac{sec ^{2} ( x ) - tan ^{2} ( x )}{sec ( x ) - tan ( x )}

使用平方差公式:

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

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

= \frac{( sec ( x ) + tan ( x ) ) ( sec ( x ) - tan ( x ) )}{sec ( x ) - tan ( x )}

约分:

sec (x) - tan (x)

= sec (x) + tan (x)

= sec (x) + tan (x)

= sec (x) + tan (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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

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

p ro v e 2 (sec (x) cot (x)) = 2 csc (x)

p ro v e sec^{2} (x) = \frac{2}{( 1 + cos ^{2} ( x ) )}

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