What is the general solution for arctan(0.5)-arctan(1/3)=arctan(x) ?

The general solution for arctan(0.5)-arctan(1/3)=arctan(x) is x= 1/7

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arctan(0.5)-arctan(1/3)=arctan(x)

Solution

arctan (0.5) - arctan (\frac{1}{3}) = arctan (x)

Solution

x = \frac{1}{7}

Solution steps

arctan (0.5) - arctan (\frac{1}{3}) = arctan (x)

arctan (0.5) = arctan (\frac{1}{2})

arctan (0.5)

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

Switch sides

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

Apply trig inverse properties

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

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

x = tan (arctan (\frac{1}{2}) - arctan (\frac{1}{3}))

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

tan (arctan (\frac{1}{2}) - arctan (\frac{1}{3}))

Rewrite using trig identities:

\frac{tan ( arctan ( \frac{1}{2} ) ) - tan ( arctan ( \frac{1}{3} ) )}{1 + tan ( arctan ( \frac{1}{2} ) ) tan ( arctan ( \frac{1}{3} ) )}

tan (arctan (\frac{1}{2}) - arctan (\frac{1}{3}))

Use the Angle Difference identity:

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

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

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

Rewrite using trig identities:

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

Use the following identity:

tan (arctan (x)) = x

= \frac{1}{2}

Rewrite using trig identities:

tan (arctan (\frac{1}{3})) = \frac{1}{3}

Use the following identity:

tan (arctan (x)) = x

= \frac{1}{3}

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

Simplify

\frac{\frac{1}{2} - \frac{1}{3}}{1 + \frac{1}{2} \cdot \frac{1}{3}} : \frac{1}{7}

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

\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{1}{6}

= \frac{\frac{1}{2} - \frac{1}{3}}{1 + \frac{1}{6}}

Join

\frac{1}{2} - \frac{1}{3} : \frac{1}{6}

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

Least Common Multiplier of

2, 3 : 6

2, 3

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

2

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6

For

\frac{1}{2} :

multiply the denominator and numerator by

3

\frac{1}{2} = \frac{1 \cdot 3}{2 \cdot 3} = \frac{3}{6}

For

\frac{1}{3} :

multiply the denominator and numerator by

2

\frac{1}{3} = \frac{1 \cdot 2}{3 \cdot 2} = \frac{2}{6}

= \frac{3}{6} - \frac{2}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{3 - 2}{6}

Subtract the numbers:

3 - 2 = 1

= \frac{1}{6}

= \frac{\frac{1}{6}}{1 + \frac{1}{6}}

Apply the fraction rule:

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

= \frac{1}{6 ( 1 + \frac{1}{6} )}

Join

1 + \frac{1}{6} : \frac{7}{6}

1 + \frac{1}{6}

Convert element to fraction:

1 = \frac{1 \cdot 6}{6}

= \frac{1 \cdot 6}{6} + \frac{1}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 6 + 1}{6}

1 \cdot 6 + 1 = 7

1 \cdot 6 + 1

Multiply the numbers:

1 \cdot 6 = 6

= 6 + 1

Add the numbers:

6 + 1 = 7

= 7

= \frac{7}{6}

= \frac{1}{6 \cdot \frac{7}{6}}

Multiply

6 \cdot \frac{7}{6} : 7

6 \cdot \frac{7}{6}

Multiply fractions:

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

= \frac{7 \cdot 6}{6}

Cancel the common factor:

6

= 7

= \frac{1}{7}

= \frac{1}{7}

x = \frac{1}{7}

x = \frac{1}{7}

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Frequently Asked Questions (FAQ)

What is the general solution for arctan(0.5)-arctan(1/3)=arctan(x) ?
The general solution for arctan(0.5)-arctan(1/3)=arctan(x) is x= 1/7