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受欢迎的三角函数 >

证明 sin^2(a)+sin^2(a)tan^2(a)=tan^2(a)

解答

证明

sin^{2} (a) + sin^{2} (a) tan^{2} (a) = tan^{2} (a)

解答

真

求解步骤

sin^{2} (a) + sin^{2} (a) tan^{2} (a) = tan^{2} (a)

调整左侧

sin^{2} (a) + sin^{2} (a) tan^{2} (a)

使用三角恒等式改写

sin^{2} (a) + sin^{2} (a) tan^{2} (a)

使用毕达哥拉斯恒等式:

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

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

= sin^{2} (a) + sin^{2} (a) (sec^{2} (a) - 1)

化简

sin^{2} (a) + sin^{2} (a) (sec^{2} (a) - 1) : sin^{2} (a) sec^{2} (a)

sin^{2} (a) + sin^{2} (a) (sec^{2} (a) - 1)

乘开

sin^{2} (a) (sec^{2} (a) - 1) : sin^{2} (a) sec^{2} (a) - sin^{2} (a)

sin^{2} (a) (sec^{2} (a) - 1)

使用分配律:

a (b - c) = ab - a c

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

= sin^{2} (a) sec^{2} (a) - sin^{2} (a) \cdot 1

= sin^{2} (a) sec^{2} (a) - 1 \cdot sin^{2} (a)

乘以:

1 \cdot sin^{2} (a) = sin^{2} (a)

= sin^{2} (a) sec^{2} (a) - sin^{2} (a)

= sin^{2} (a) + sin^{2} (a) sec^{2} (a) - sin^{2} (a)

同类项相加:

sin^{2} (a) - sin^{2} (a) = 0

= sin^{2} (a) sec^{2} (a)

= sin^{2} (a) sec^{2} (a)

= sin^{2} (a) sec^{2} (a)

用 sin, cos 表示

sec^{2} (a) sin^{2} (a)

使用基本三角恒等式:

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

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

化简

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

(\frac{1}{cos ( a )})^{2} sin^{2} (a)

(\frac{1}{cos ( a )})^{2} = \frac{1}{cos ^{2} ( a )}

(\frac{1}{cos ( a )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( a )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( a )}

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

分式相乘:

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

= \frac{1 \cdot sin ^{2} ( a )}{cos ^{2} ( a )}

乘以:

1 \cdot sin^{2} (a) = sin^{2} (a)

= \frac{sin ^{2} ( a )}{cos ^{2} ( a )}

= \frac{sin ^{2} ( a )}{cos ^{2} ( a )}

= \frac{sin ^{2} ( a )}{cos ^{2} ( a )}

使用三角恒等式改写

= \frac{sin ( a )}{cos ( a )} \cdot \frac{sin ( a )}{cos ( a )}

使用基本三角恒等式:

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

\frac{sin ( a ) tan ( a )}{cos ( a )}

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

使用基本三角恒等式:

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

tan (a) tan (a)

化简

tan (a) tan (a) : tan^{2} (a)

tan (a) tan (a)

使用指数法则:

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

tan (a) tan (a) = tan^{1 + 1} (a)

= tan^{1 + 1} (a)

数字相加:

1 + 1 = 2

= tan^{2} (a)

tan^{2} (a)

tan^{2} (a)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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