What is the general solution for tan(θ)=(-11/3-(-4/3))/(1+(-4/3)(-11/3)) ?

The general solution for tan(θ)=(-11/3-(-4/3))/(1+(-4/3)(-11/3)) is θ=-0.37724…+pin

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Popular >

tan(θ)=(-11/3-(-4/3))/(1+(-4/3)(-11/3))

Solution

tan (θ) = \frac{- \frac{11}{3} - ( - \frac{4}{3} )}{1 + ( - \frac{4}{3} ) ( - \frac{11}{3} )}

Solution

θ = - 0.37724 \dots + πn

Solution steps

tan (θ) = \frac{- \frac{11}{3} - ( - \frac{4}{3} )}{1 + ( - \frac{4}{3} ) ( - \frac{11}{3} )}

Simplify

\frac{- \frac{11}{3} - ( - \frac{4}{3} )}{1 + ( - \frac{4}{3} ) ( - \frac{11}{3} )} : - \frac{21}{53}

\frac{- \frac{11}{3} - ( - \frac{4}{3} )}{1 + ( - \frac{4}{3} ) ( - \frac{11}{3} )}

Remove parentheses:

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

= \frac{- \frac{11}{3} + \frac{4}{3}}{1 + \frac{4}{3} \cdot \frac{11}{3}}

\frac{4}{3} \cdot \frac{11}{3} = \frac{44}{9}

\frac{4}{3} \cdot \frac{11}{3}

Multiply fractions:

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

= \frac{4 \cdot 11}{3 \cdot 3}

Multiply the numbers:

4 \cdot 11 = 44

= \frac{44}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{44}{9}

= \frac{- \frac{11}{3} + \frac{4}{3}}{1 + \frac{44}{9}}

Join

- \frac{11}{3} + \frac{4}{3} : - \frac{7}{3}

- \frac{11}{3} + \frac{4}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{- 11 + 4}{3}

Add/Subtract the numbers:

- 11 + 4 = - 7

= \frac{- 7}{3}

Apply the fraction rule:

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

= - \frac{7}{3}

= \frac{- \frac{7}{3}}{1 + \frac{44}{9}}

Apply the fraction rule:

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

= - \frac{\frac{7}{3}}{1 + \frac{44}{9}}

Apply the fraction rule:

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

\frac{\frac{7}{3}}{1 + \frac{44}{9}} = \frac{7}{3 ( 1 + \frac{44}{9} )}

= - \frac{7}{3 ( 1 + \frac{44}{9} )}

Join

1 + \frac{44}{9} : \frac{53}{9}

1 + \frac{44}{9}

Convert element to fraction:

1 = \frac{1 \cdot 9}{9}

= \frac{1 \cdot 9}{9} + \frac{44}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 9 + 44}{9}

1 \cdot 9 + 44 = 53

1 \cdot 9 + 44

Multiply the numbers:

1 \cdot 9 = 9

= 9 + 44

Add the numbers:

9 + 44 = 53

= 53

= \frac{53}{9}

= - \frac{7}{3 \cdot \frac{53}{9}}

Multiply

3 \cdot \frac{53}{9} : \frac{53}{3}

3 \cdot \frac{53}{9}

Multiply fractions:

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

= \frac{53 \cdot 3}{9}

Multiply the numbers:

53 \cdot 3 = 159

= \frac{159}{9}

Cancel the common factor:

3

= \frac{53}{3}

= - \frac{7}{\frac{53}{3}}

Simplify

\frac{7}{\frac{53}{3}} : \frac{21}{53}

\frac{7}{\frac{53}{3}}

Apply the fraction rule:

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

= \frac{7 \cdot 3}{53}

Multiply the numbers:

7 \cdot 3 = 21

= \frac{21}{53}

= - \frac{21}{53}

tan (θ) = - \frac{21}{53}

Apply trig inverse properties

tan (θ) = - \frac{21}{53}

General solutions for

tan (θ) = - \frac{21}{53}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- \frac{21}{53}) + πn

θ = arctan (- \frac{21}{53}) + πn

Show solutions in decimal form

θ = - 0.37724 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(θ)=(-11/3-(-4/3))/(1+(-4/3)(-11/3)) ?
The general solution for tan(θ)=(-11/3-(-4/3))/(1+(-4/3)(-11/3)) is θ=-0.37724…+pin