证明 tan^2(x)-(sec(x)+1)(sec(x)-1)=0

解答

证明

tan^{2} (x) - (sec (x) + 1) (sec (x) - 1) = 0

真

求解步骤

tan^{2} (x) - (sec (x) + 1) (sec (x) - 1) = 0

调整左侧

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

使用三角恒等式改写

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

乘开

(sec (x) + 1) (sec (x) - 1) : sec^{2} (x) - 1

(sec (x) + 1) (sec (x) - 1)

使用平方差公式:

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

a = sec (x), b = 1

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

使用法则

1^{a} = 1

1^{2} = 1

= sec^{2} (x) - 1

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

使用毕达哥拉斯恒等式:

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

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

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

p ro v e sec^{4} (θ) = sec^{2} (θ) tan^{2} (θ) + sec^{2} (θ)

p ro v e sin^{2} (θ) = 1 - \frac{1}{sec ^{2} ( θ )}

p ro v e tan (θ) sec (θ) csc (θ) = 1 + tan^{2} (θ)

p ro v e \frac{1 - cos ( x )}{\frac{1}{c o s ( x )} - 1} = cos (x)

p ro v e sin (π - 0) = sin (0)