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受欢迎的三角函数 >

证明 tan^4(t)+tan^2(t)=sec^4(t)-sec^2(t)

解答

证明

tan^{4} (t) + tan^{2} (t) = sec^{4} (t) - sec^{2} (t)

解答

真

求解步骤

tan^{4} (t) + tan^{2} (t) = sec^{4} (t) - sec^{2} (t)

调整左侧

tan^{4} (t) + tan^{2} (t)

分解

tan^{2} (t) + tan^{4} (t) : tan^{2} (t) (1 + tan^{2} (t))

tan^{2} (t) + tan^{4} (t)

使用指数法则:

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

tan^{4} (t) = tan^{2} (t) tan^{2} (t)

= tan^{2} (t) + tan^{2} (t) tan^{2} (t)

因式分解出通项

tan^{2} (t)

= tan^{2} (t) (1 + tan^{2} (t))

= (1 + tan^{2} (t)) tan^{2} (t)

使用三角恒等式改写

(1 + tan^{2} (t)) tan^{2} (t)

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

对同类项分组

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

1 - 1 = 0

= sec^{2} (t)

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

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

乘开

(- 1 + sec^{2} (t)) sec^{2} (t) : - sec^{2} (t) + sec^{4} (t)

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

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

使用分配律:

a (b + c) = ab + a c

a = sec^{2} (t), b = - 1, c = sec^{2} (t)

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

使用加减运算法则

+ (- a) = - a

= - 1 \cdot sec^{2} (t) + sec^{2} (t) sec^{2} (t)

化简

- 1 \cdot sec^{2} (t) + sec^{2} (t) sec^{2} (t) : - sec^{2} (t) + sec^{4} (t)

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

1 \cdot sec^{2} (t) = sec^{2} (t)

1 \cdot sec^{2} (t)

乘以:

1 \cdot sec^{2} (t) = sec^{2} (t)

= sec^{2} (t)

sec^{2} (t) sec^{2} (t) = sec^{4} (t)

sec^{2} (t) sec^{2} (t)

使用指数法则:

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

sec^{2} (t) sec^{2} (t) = sec^{2 + 2} (t)

= sec^{2 + 2} (t)

数字相加:

2 + 2 = 4

= sec^{4} (t)

= - sec^{2} (t) + sec^{4} (t)

= - sec^{2} (t) + sec^{4} (t)

= - sec^{2} (t) + sec^{4} (t)

= sec^{4} (t) - sec^{2} (t)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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