What is the general solution for arctan(x)+arctan(1-x)=arctan(9/7) ?

The general solution for arctan(x)+arctan(1-x)=arctan(9/7) is x= 2/3 ,x= 1/3

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arctan(x)+arctan(1-x)=arctan(9/7)

Solution

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

Solution

x = \frac{2}{3}, x = \frac{1}{3}

Solution steps

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

Rewrite using trig identities

arctan (x) + arctan (1 - x)

Use the Sum to Product identity:

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

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

arctan (\frac{x + 1 - x}{1 - x ( 1 - x )}) = arctan (\frac{9}{7})

Apply trig inverse properties

arctan (\frac{x + 1 - x}{1 - x ( 1 - x )}) = arctan (\frac{9}{7})

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

\frac{x + 1 - x}{1 - x ( 1 - x )} = tan (arctan (\frac{9}{7}))

tan (arctan (\frac{9}{7})) = \frac{9}{7}

tan (arctan (\frac{9}{7}))

Rewrite using trig identities:

tan (arctan (\frac{9}{7})) = \frac{9}{7}

Use the following identity:

tan (arctan (x)) = x

= \frac{9}{7}

= \frac{9}{7}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7}

Solve

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7} : x = \frac{2}{3}, x = \frac{1}{3}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7}

Cross multiply

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7}

Simplify

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

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

x + 1 - x = 1

x + 1 - x

Group like terms

= x - x + 1

Add similar elements:

x - x = 0

= 1

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

\frac{1}{1 - x ( 1 - x )} = \frac{9}{7}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

1 \cdot 7 = (1 - x (1 - x)) \cdot 9

Simplify

1 \cdot 7 : 7

1 \cdot 7

Multiply the numbers:

1 \cdot 7 = 7

= 7

7 = (1 - x (1 - x)) \cdot 9

7 = (1 - x (1 - x)) \cdot 9

Solve

7 = (1 - x (1 - x)) \cdot 9 : x = \frac{2}{3}, x = \frac{1}{3}

7 = (1 - x (1 - x)) \cdot 9

Expand

(1 - x (1 - x)) \cdot 9 : 9 - 9 x + 9 x^{2}

(1 - x (1 - x)) \cdot 9

Expand

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

1 - x (1 - x)

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 - (- x) x

Apply minus-plus rules

- (- a) = a

= - 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}

= 1 - x + x^{2}

= 9 (x^{2} - x + 1)

= 9 (1 - x + x^{2})

Distribute parentheses

= 9 \cdot 1 + 9 (- x) + 9 x^{2}

Apply minus-plus rules

+ (- a) = - a

= 9 \cdot 1 - 9 x + 9 x^{2}

Multiply the numbers:

9 \cdot 1 = 9

= 9 - 9 x + 9 x^{2}

7 = 9 - 9 x + 9 x^{2}

Switch sides

9 - 9 x + 9 x^{2} = 7

Move

7

to the left side

9 - 9 x + 9 x^{2} = 7

Subtract

7

from both sides

9 - 9 x + 9 x^{2} - 7 = 7 - 7

Simplify

9 x^{2} - 9 x + 2 = 0

9 x^{2} - 9 x + 2 = 0

Solve with the quadratic formula

9 x^{2} - 9 x + 2 = 0

Quadratic Equation Formula:

For

a = 9, b = - 9, c = 2

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 9 \cdot 2}{2 \cdot 9}

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 9 \cdot 2}{2 \cdot 9}

(- 9)^{2} - 4 \cdot 9 \cdot 2 = 3

(- 9)^{2} - 4 \cdot 9 \cdot 2

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 9 \cdot 2

Multiply the numbers:

4 \cdot 9 \cdot 2 = 72

= 9^{2} - 72

9^{2} = 81

= 81 - 72

Subtract the numbers:

81 - 72 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

x_{1, 2} = \frac{- ( - 9 ) \pm 3}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 3}{2 \cdot 9}, x_{2} = \frac{- ( - 9 ) - 3}{2 \cdot 9}

x = \frac{- ( - 9 ) + 3}{2 \cdot 9} : \frac{2}{3}

\frac{- ( - 9 ) + 3}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{9 + 3}{2 \cdot 9}

Add the numbers:

9 + 3 = 12

= \frac{12}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{12}{18}

Cancel the common factor:

6

= \frac{2}{3}

x = \frac{- ( - 9 ) - 3}{2 \cdot 9} : \frac{1}{3}

\frac{- ( - 9 ) - 3}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{9 - 3}{2 \cdot 9}

Subtract the numbers:

9 - 3 = 6

= \frac{6}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{6}{18}

Cancel the common factor:

6

= \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = \frac{1}{3}

x = \frac{2}{3}, x = \frac{1}{3}

x = \frac{2}{3}, x = \frac{1}{3}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

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

Check the solution

\frac{2}{3} :

True

\frac{2}{3}

Plug in

n = 1

\frac{2}{3}

For

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

plug in

x = \frac{2}{3}

arctan (\frac{2}{3}) + arctan (1 - \frac{2}{3}) = arctan (\frac{9}{7})

Refine

0.90975 \dots = 0.90975 \dots

\Rightarrow True

Check the solution

\frac{1}{3} :

True

\frac{1}{3}

Plug in

n = 1

\frac{1}{3}

For

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

plug in

x = \frac{1}{3}

arctan (\frac{1}{3}) + arctan (1 - \frac{1}{3}) = arctan (\frac{9}{7})

Refine

0.90975 \dots = 0.90975 \dots

\Rightarrow True

x = \frac{2}{3}, x = \frac{1}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for arctan(x)+arctan(1-x)=arctan(9/7) ?
The general solution for arctan(x)+arctan(1-x)=arctan(9/7) is x= 2/3 ,x= 1/3