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

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

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Popular Trigonometry >

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

Solution

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

Solution

x = 2, x = - 2

Solution steps

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

Rewrite using trig identities

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

Use the Sum to Product identity:

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

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

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

Apply trig inverse properties

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

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

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

tan (arctan (\frac{1}{2})) = \frac{1}{2}

tan (arctan (\frac{1}{2}))

Rewrite using trig identities:

tan (arctan (\frac{1}{2})) = \frac{1}{2}

Use the following identity:

tan (arctan (x)) = x

= \frac{1}{2}

= \frac{1}{2}

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

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

Solve

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

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

Cross multiply

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

Simplify

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

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

1 + x + 1 - x = 2

1 + x + 1 - x

Group like terms

= x - x + 1 + 1

Add similar elements:

x - x = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

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

Expand

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

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

Expand

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

Expand

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

(1 + x) (1 - x)

Apply Difference of Two Squares Formula:

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

a = 1, b = x

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - x^{2}

= - (1 - x^{2})

Distribute parentheses

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

Apply minus-plus rules

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

= - 1 + x^{2}

= 1 - 1 + x^{2}

1 - 1 = 0

= x^{2}

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

\frac{2}{x ^{2}} = \frac{1}{2}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

2 \cdot 2 = x^{2} \cdot 1

Simplify

2 \cdot 2 = x^{2} \cdot 1

Simplify

2 \cdot 2 : 4

2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Simplify

x^{2} \cdot 1 : x^{2}

x^{2} \cdot 1

Multiply:

x^{2} \cdot 1 = x^{2}

= x^{2}

4 = x^{2}

4 = x^{2}

4 = x^{2}

Solve

4 = x^{2} : x = 2, x = - 2

4 = x^{2}

Switch sides

x^{2} = 4

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 4, x = - 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

- 4 = - 2

- 4

Factor the number:

4 = 2^{2}

= - 2^{2}

Apply radical rule:

a^{2} = a, a \geq 0

2^{2} = - 2

= - 2

x = 2, x = - 2

x = 2, x = - 2

Verify Solutions

Find undefined (singularity) points:

x = 0

Take the denominator(s) of

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

and compare to zero

Solve

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

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

Expand

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

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

Expand

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

Expand

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

(1 + x) (1 - x)

Apply Difference of Two Squares Formula:

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

a = 1, b = x

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - x^{2}

= - (1 - x^{2})

Distribute parentheses

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

Apply minus-plus rules

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

= - 1 + x^{2}

= 1 - 1 + x^{2}

1 - 1 = 0

= x^{2}

x^{2} = 0

Solve with the quadratic formula

x^{2} = 0

Quadratic Equation Formula:

For

a = 1, b = 0, c = 0

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

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

0^{2} - 4 \cdot 1 \cdot 0 = 0

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

Apply rule

0^{a} = 0

0^{2} = 0

= 0 - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0 - 0

Subtract the numbers:

0 - 0 = 0

= 0

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

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

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

x = 0

The solution to the quadratic equation is:

x = 0

The following points are undefined

x = 0

Combine undefined points with solutions:

x = 2, x = - 2

x = 2, x = - 2

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

2 :

True

2

Plug in

n = 1

2

For

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

plug in

x = 2

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

Refine

0.46364 \dots = 0.46364 \dots

\Rightarrow True

Check the solution

- 2 :

True

- 2

Plug in

n = 1

- 2

For

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

plug in

x = - 2

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

Refine

0.46364 \dots = 0.46364 \dots

\Rightarrow True

x = 2, x = - 2

Graph

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Frequently Asked Questions (FAQ)

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