What is the general solution for arctan(x+1)+arctan(x-1)=arctan(2) ?

The general solution for arctan(x+1)+arctan(x-1)=arctan(2) is x=1

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arctan(x+1)+arctan(x-1)=arctan(2)

Solution

arctan (x + 1) + arctan (x - 1) = arctan (2)

Solution

x = 1

Solution steps

arctan (x + 1) + arctan (x - 1) = arctan (2)

Rewrite using trig identities

arctan (x + 1) + arctan (x - 1)

Use the Sum to Product identity:

arctan (s) + arctan (t) = arctan (\frac{s + t}{1 - s t})

= arctan (\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )})

arctan (\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )}) = arctan (2)

Apply trig inverse properties

arctan (\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )}) = arctan (2)

arctan (x) = a \Rightarrow x = tan (a)

\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )} = tan (arctan (2))

tan (arctan (2)) = 2

tan (arctan (2))

Rewrite using trig identities:

tan (arctan (2)) = 2

Use the following identity:

tan (arctan (x)) = x

= 2

= 2

\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )} = 2

\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )} = 2

Solve

\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )} = 2 : x = - 2, x = 1

\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )} = 2

Simplify

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

\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )}

x + 1 + x - 1 = 2 x

x + 1 + x - 1

Group like terms

= x + x + 1 - 1

Add similar elements:

x + x = 2 x

= 2 x + 1 - 1

1 - 1 = 0

= 2 x

= \frac{2 x}{1 - ( x + 1 ) ( x - 1 )}

Expand

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

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

Expand

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

Expand

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

(x + 1) (x - 1)

Apply Difference of Two Squares Formula:

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

a = x, b = 1

= x^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= x^{2} - 1

= - (x^{2} - 1)

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - x^{2} + 1

= 1 - x^{2} + 1

Simplify

1 - x^{2} + 1 : - x^{2} + 2

1 - x^{2} + 1

Group like terms

= - x^{2} + 1 + 1

Add the numbers:

1 + 1 = 2

= - x^{2} + 2

= - x^{2} + 2

= \frac{2 x}{- x ^{2} + 2}

\frac{2 x}{- x ^{2} + 2} = 2

Multiply both sides by

- x^{2} + 2

\frac{2 x}{- x ^{2} + 2} = 2

Multiply both sides by

- x^{2} + 2

\frac{2 x}{- x ^{2} + 2} (- x^{2} + 2) = 2 (- x^{2} + 2)

Simplify

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

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

Solve

2 x = 2 (- x^{2} + 2) : x = - 2, x = 1

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

Expand

2 (- x^{2} + 2) : - 2 x^{2} + 4

2 (- x^{2} + 2)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = - x^{2}, c = 2

= 2 (- x^{2}) + 2 \cdot 2

Apply minus-plus rules

+ (- a) = - a

= - 2 x^{2} + 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= - 2 x^{2} + 4

2 x = - 2 x^{2} + 4

Switch sides

- 2 x^{2} + 4 = 2 x

Move

2 x

to the left side

- 2 x^{2} + 4 = 2 x

Subtract

2 x

from both sides

- 2 x^{2} + 4 - 2 x = 2 x - 2 x

Simplify

- 2 x^{2} + 4 - 2 x = 0

- 2 x^{2} + 4 - 2 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 x^{2} - 2 x + 4 = 0

Solve with the quadratic formula

- 2 x^{2} - 2 x + 4 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 2, c = 4

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) \cdot 4}{2 ( - 2 )}

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) \cdot 4}{2 ( - 2 )}

(- 2)^{2} - 4 (- 2) \cdot 4 = 6

(- 2)^{2} - 4 (- 2) \cdot 4

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 2 \cdot 4

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 2^{2} + 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

x_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 6}{2 ( - 2 )}, x_{2} = \frac{- ( - 2 ) - 6}{2 ( - 2 )}

x = \frac{- ( - 2 ) + 6}{2 ( - 2 )} : - 2

\frac{- ( - 2 ) + 6}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{2 + 6}{- 2 \cdot 2}

Add the numbers:

2 + 6 = 8

= \frac{8}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

x = \frac{- ( - 2 ) - 6}{2 ( - 2 )} : 1

\frac{- ( - 2 ) - 6}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{2 - 6}{- 2 \cdot 2}

Subtract the numbers:

2 - 6 = - 4

= \frac{- 4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

x = - 2, x = 1

x = - 2, x = 1

Verify Solutions

Find undefined (singularity) points:

x = - 2, x = 2

Take the denominator(s) of

\frac{x + 1 + x - 1}{1 - ( x + 1 ) ( x - 1 )}

and compare to zero

Solve

1 - (x + 1) (x - 1) = 0 : x = - 2, x = 2

1 - (x + 1) (x - 1) = 0

Expand

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

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

Expand

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

Expand

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

(x + 1) (x - 1)

Apply Difference of Two Squares Formula:

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

a = x, b = 1

= x^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= x^{2} - 1

= - (x^{2} - 1)

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - x^{2} + 1

= 1 - x^{2} + 1

Simplify

1 - x^{2} + 1 : - x^{2} + 2

1 - x^{2} + 1

Group like terms

= - x^{2} + 1 + 1

Add the numbers:

1 + 1 = 2

= - x^{2} + 2

= - x^{2} + 2

- x^{2} + 2 = 0

Solve with the quadratic formula

- x^{2} + 2 = 0

Quadratic Equation Formula:

For

a = - 1, b = 0, c = 2

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \cdot 2}{2 ( - 1 )}

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \cdot 2}{2 ( - 1 )}

0^{2} - 4 (- 1) \cdot 2 = 22

0^{2} - 4 (- 1) \cdot 2

Apply rule

0^{a} = 0

0^{2} = 0

= 0 - 4 (- 1) \cdot 2

Apply rule

- (- a) = a

= 0 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 0 + 8

Add the numbers:

0 + 8 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

x_{1, 2} = \frac{- 0 \pm 2 2}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 0 + 2 2}{2 ( - 1 )}, x_{2} = \frac{- 0 - 2 2}{2 ( - 1 )}

x = \frac{- 0 + 2 2}{2 ( - 1 )} : - 2

\frac{- 0 + 2 2}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 0 + 2 2}{- 2 \cdot 1}

- 0 + 22 = 22

= \frac{2 2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 2}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2 2}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 2

x = \frac{- 0 - 2 2}{2 ( - 1 )} : 2

\frac{- 0 - 2 2}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 0 - 2 2}{- 2 \cdot 1}

- 0 - 22 = - 22

= \frac{- 2 2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 2}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2

The solutions to the quadratic equation are:

x = - 2, x = 2

The following points are undefined

x = - 2, x = 2

Combine undefined points with solutions:

x = - 2, x = 1

x = - 2, x = 1

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arctan (x + 1) + arctan (x - 1) = arctan (2)

Remove the ones that don't agree with the equation.

Check the solution

- 2 :

False

- 2

Plug in

n = 1

- 2

For

arctan (x + 1) + arctan (x - 1) = arctan (2)

plug in

x = - 2

arctan (- 2 + 1) + arctan (- 2 - 1) = arctan (2)

Refine

- 2.03444 \dots = 1.10714 \dots

\Rightarrow False

Check the solution

1 :

True

1

Plug in

n = 1

1

For

arctan (x + 1) + arctan (x - 1) = arctan (2)

plug in

x = 1

arctan (1 + 1) + arctan (1 - 1) = arctan (2)

Refine

1.10714 \dots = 1.10714 \dots

\Rightarrow True

x = 1

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Frequently Asked Questions (FAQ)

What is the general solution for arctan(x+1)+arctan(x-1)=arctan(2) ?
The general solution for arctan(x+1)+arctan(x-1)=arctan(2) is x=1