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受欢迎的三角函数 >

(1-tan^2(A))/(1+tan^2(A))=1

解答

\frac{1 - tan ^{2} ( A )}{1 + tan ^{2} ( A )} = 1

解答

A = πn

+1

度数

A = 0^{\circ} + 18 0^{\circ} n

求解步骤

\frac{1 - tan ^{2} ( A )}{1 + tan ^{2} ( A )} = 1

用替代法求解

\frac{1 - tan ^{2} ( A )}{1 + tan ^{2} ( A )} = 1

令

: tan (A) = u

\frac{1 - u ^{2}}{1 + u ^{2}} = 1

\frac{1 - u ^{2}}{1 + u ^{2}} = 1 : u = 0

\frac{1 - u ^{2}}{1 + u ^{2}} = 1

在两边乘以

1 + u^{2}

\frac{1 - u ^{2}}{1 + u ^{2}} = 1

在两边乘以

1 + u^{2}

\frac{1 - u ^{2}}{1 + u ^{2}} (1 + u^{2}) = 1 \cdot (1 + u^{2})

化简

\frac{1 - u ^{2}}{1 + u ^{2}} (1 + u^{2}) = 1 \cdot (1 + u^{2})

化简

\frac{1 - u ^{2}}{1 + u ^{2}} (1 + u^{2}) : 1 - u^{2}

\frac{1 - u ^{2}}{1 + u ^{2}} (1 + u^{2})

分式相乘:

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

= \frac{( 1 - u ^{2} ) ( 1 + u ^{2} )}{1 + u ^{2}}

约分:

1 + u^{2}

= 1 - u^{2}

化简

1 \cdot (1 + u^{2}) : 1 + u^{2}

1 \cdot (1 + u^{2})

乘以:

1 \cdot (1 + u^{2}) = (1 + u^{2})

= (1 + u^{2})

去除括号:

(a) = a

= 1 + u^{2}

1 - u^{2} = 1 + u^{2}

1 - u^{2} = 1 + u^{2}

1 - u^{2} = 1 + u^{2}

解

1 - u^{2} = 1 + u^{2} : u = 0

1 - u^{2} = 1 + u^{2}

将

1

到右边

1 - u^{2} = 1 + u^{2}

两边减去

1

1 - u^{2} - 1 = 1 + u^{2} - 1

化简

- u^{2} = u^{2}

- u^{2} = u^{2}

将

u^{2}

para o lado esquerdo

- u^{2} = u^{2}

两边减去

u^{2}

- u^{2} - u^{2} = u^{2} - u^{2}

化简

- 2 u^{2} = 0

- 2 u^{2} = 0

两边除以

- 2

- 2 u^{2} = 0

两边除以

- 2

- 2 u^{2} = 0

两边除以

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{0}{- 2}

化简

u^{2} = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

u = 0

u = tan (A)

代回

tan (A) = 0

tan (A) = 0

tan (A) = 0 : A = πn

tan (A) = 0

tan (A) = 0

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

A = 0 + πn

A = 0 + πn

解

A = 0 + πn : A = πn

A = 0 + πn

0 + πn = πn

A = πn

A = πn

合并所有解

A = πn

作图

流行的例子

sin (2 x) - 0.8 = 0

tan(a)=sqrt(15/7),sin(a)

tan (a) = \frac{15}{7}, sin (a)

2cos^2(θ)+sin(θ)=2

2 cos^{2} (θ) + sin (θ) = 2

tanh^2(x)+5sech(x)-5=0

tanh^{2} (x) + 5 sech (x) - 5 = 0

solvefor a,P=cot^2(a)

so l v e f or a, P = cot^{2} (a)