What is the general solution for arccot(x)+arccot(1+x)= pi/4 ?

The general solution for arccot(x)+arccot(1+x)= pi/4 is x=2

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arccot(x)+arccot(1+x)= pi/4

Solution

arccot (x) + arccot (1 + x) = \frac{π}{4}

Solution

x = 2

Solution steps

arccot (x) + arccot (1 + x) = \frac{π}{4}

Rewrite using trig identities

arccot (x) + arccot (1 + x)

Use the Sum to Product identity:

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

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

arccot (\frac{x ( 1 + x ) - 1}{1 + x + x}) = \frac{π}{4}

Apply trig inverse properties

arccot (\frac{x ( 1 + x ) - 1}{1 + x + x}) = \frac{π}{4}

arccot (x) = a \Rightarrow x = cot (a)

\frac{x ( 1 + x ) - 1}{1 + x + x} = cot (\frac{π}{4})

cot (\frac{π}{4}) = 1

cot (\frac{π}{4})

Use the following trivial identity:

cot (\frac{π}{4}) = 1

cot (\frac{π}{4})

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= 1

= 1

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

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

Solve

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

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

Simplify

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

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

Add similar elements:

x + x = 2 x

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

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

Multiply both sides by

1 + 2 x

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

Multiply both sides by

1 + 2 x

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

1 + 2 x

= x (1 + x) - 1

Simplify

1 \cdot (1 + 2 x) : 1 + 2 x

1 \cdot (1 + 2 x)

Multiply:

1 \cdot (1 + 2 x) = (1 + 2 x)

= (1 + 2 x)

Remove parentheses:

(a) = a

= 1 + 2 x

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

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

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

Solve

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

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

Expand

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

x (1 + x) - 1

Expand

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

x (1 + x)

Apply the distributive law:

a (b + c) = ab + a c

a = x, b = 1, c = x

= x \cdot 1 + xx

= 1 \cdot x + xx

Simplify

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

1 \cdot x + xx

1 \cdot x = x

1 \cdot x

Multiply:

1 \cdot x = x

= x

xx = x^{2}

xx

Apply exponent rule:

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

xx = x^{1 + 1}

= x^{1 + 1}

Add the numbers:

1 + 1 = 2

= x^{2}

= x + x^{2}

= x + x^{2}

= x + x^{2} - 1

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

Move

2 x

to the left side

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

Subtract

2 x

from both sides

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

Simplify

x^{2} - x - 1 = 1

x^{2} - x - 1 = 1

Move

1

to the left side

x^{2} - x - 1 = 1

Subtract

1

from both sides

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

Simplify

x^{2} - x - 2 = 0

x^{2} - x - 2 = 0

Solve with the quadratic formula

x^{2} - x - 2 = 0

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 2) = 3

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

x_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 1}

x = \frac{- ( - 1 ) + 3}{2 \cdot 1} : 2

\frac{- ( - 1 ) + 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 + 3}{2 \cdot 1}

Add the numbers:

1 + 3 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

x = \frac{- ( - 1 ) - 3}{2 \cdot 1} : - 1

\frac{- ( - 1 ) - 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 - 3}{2 \cdot 1}

Subtract the numbers:

1 - 3 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

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 = - \frac{1}{2}

Take the denominator(s) of

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

and compare to zero

Solve

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

1 + x + x = 0

Add similar elements:

x + x = 2 x

1 + 2 x = 0

Move

1

to the right side

1 + 2 x = 0

Subtract

1

from both sides

1 + 2 x - 1 = 0 - 1

Simplify

2 x = - 1

2 x = - 1

Divide both sides by

2

2 x = - 1

Divide both sides by

2

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

Simplify

x = - \frac{1}{2}

x = - \frac{1}{2}

The following points are undefined

x = - \frac{1}{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

arccot (x) + arccot (1 + x) = \frac{π}{4}

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

Check the solution

2 :

True

2

Plug in

n = 1

2

For

arccot (x) + arccot (1 + x) = \frac{π}{4}

plug in

x = 2

arccot (2) + arccot (1 + 2) = \frac{π}{4}

Refine

0.78539 \dots = 0.78539 \dots

\Rightarrow True

Check the solution

- 1 :

False

- 1

Plug in

n = 1

- 1

For

arccot (x) + arccot (1 + x) = \frac{π}{4}

plug in

x = - 1

arccot (- 1) + arccot (1 - 1) = \frac{π}{4}

Refine

3.92699 \dots = 0.78539 \dots

\Rightarrow False

x = 2

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Frequently Asked Questions (FAQ)

What is the general solution for arccot(x)+arccot(1+x)= pi/4 ?
The general solution for arccot(x)+arccot(1+x)= pi/4 is x=2