What is the general solution for (1-tan^2(A))/(1+tan^2(A))=1 ?

The general solution for (1-tan^2(A))/(1+tan^2(A))=1 is A=pin

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

Solution

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

Solution

A = πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

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

Multiply both sides by

1 + u^{2}

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

Multiply both sides by

1 + u^{2}

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

1 + u^{2}

= 1 - u^{2}

Simplify

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

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

Multiply:

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

= (1 + u^{2})

Remove parentheses:

(a) = a

= 1 + u^{2}

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

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

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

Solve

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

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

Move

u^{2}

to the left side

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

Subtract

u^{2}

from both sides

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

Simplify

- 2 u^{2} = 0

- 2 u^{2} = 0

Divide both sides by

- 2

- 2 u^{2} = 0

Divide both sides by

- 2

- 2 u^{2} = 0

Divide both sides by

- 2

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

Simplify

u^{2} = 0

u^{2} = 0

Apply rule

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

u = 0

u = 0

Substitute back

u = tan (A)

tan (A) = 0

tan (A) = 0

tan (A) = 0 : A = πn

tan (A) = 0

General solutions for

tan (A) = 0

tan (x)

periodicity table with

πn

cycle:

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

Solve

A = 0 + πn : A = πn

A = 0 + πn

0 + πn = πn

A = πn

A = πn

Combine all the solutions

A = πn

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Frequently Asked Questions (FAQ)

What is the general solution for (1-tan^2(A))/(1+tan^2(A))=1 ?
The general solution for (1-tan^2(A))/(1+tan^2(A))=1 is A=pin