What is the value of (tan(120)+tan(45))/(1-tan(120)tan(45)) ?

The value of (tan(120)+tan(45))/(1-tan(120)tan(45)) is -2+sqrt(3)

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(tan(120)+tan(45))/(1-tan(120)tan(45))

Solution

\frac{tan ( 12 0 ^{\circ} ) + tan ( 4 5 ^{\circ} )}{1 - tan ( 12 0 ^{\circ} ) tan ( 4 5 ^{\circ} )}

Solution

- 2 + 3

Decimal Notation

- 0.26794 \dots

Solution steps

\frac{tan ( 12 0 ^{\circ} ) + tan ( 4 5 ^{\circ} )}{1 - tan ( 12 0 ^{\circ} ) tan ( 4 5 ^{\circ} )}

Rewrite using trig identities:

tan (12 0^{\circ}) = - 3

tan (12 0^{\circ})

Rewrite using trig identities:

\frac{sin ( 12 0 ^{\circ} )}{cos ( 12 0 ^{\circ} )}

tan (12 0^{\circ})

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( 12 0 ^{\circ} )}{cos ( 12 0 ^{\circ} )}

= \frac{sin ( 12 0 ^{\circ} )}{cos ( 12 0 ^{\circ} )}

Use the following trivial identity:

sin (12 0^{\circ}) = \frac{3}{2}

sin (12 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

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

Simplify

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

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

Apply the fraction rule:

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

2

= - 3

= - 3

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

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

Simplify

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

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

Apply rule

- (- a) = a

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

Multiply:

3 \cdot 1 = 3

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

Rationalize

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

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

Multiply by the conjugate

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

= \frac{( - 3 + 1 ) ( 1 - 3 )}{( 1 + 3 ) ( 1 - 3 )}

(- 3 + 1) (1 - 3) = 4 - 23

(- 3 + 1) (1 - 3)

Apply exponent rule:

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

(- 3 + 1) (1 - 3) = (- 3 + 1)^{1 + 1}

= (- 3 + 1)^{1 + 1}

Add the numbers:

1 + 1 = 2

= (- 3 + 1)^{2}

Apply Perfect Square Formula:

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

a = - 3, b = 1

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

Simplify

(- 3)^{2} + 2 (- 3) \cdot 1 + 1^{2} : 4 - 23

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Remove parentheses:

(- a) = - a

= (- 3)^{2} - 23 \cdot 1 + 1

(- 3)^{2} = 3

(- 3)^{2}

Apply exponent rule:

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

n

is even

(- 3)^{2} = (3)^{2}

= (3)^{2}

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 3

23 \cdot 1 = 23

23 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23

= 3 - 23 + 1

Add the numbers:

3 + 1 = 4

= 4 - 23

= 4 - 23

(1 + 3) (1 - 3) = - 2

(1 + 3) (1 - 3)

Apply Difference of Two Squares Formula:

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

a = 1, b = 3

= 1^{2} - (3)^{2}

Simplify

1^{2} - (3)^{2} : - 2

1^{2} - (3)^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - (3)^{2}

(3)^{2} = 3

(3)^{2}

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 3

= 1 - 3

Subtract the numbers:

1 - 3 = - 2

= - 2

= - 2

= \frac{4 - 2 3}{- 2}

Apply the fraction rule:

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

= - \frac{4 - 2 3}{2}

Cancel

\frac{4 - 2 3}{2} : 2 - 3

\frac{4 - 2 3}{2}

Factor

4 - 23 : 2 (2 - 3)

4 - 23

Rewrite as

= 2 \cdot 2 - 23

Factor out common term

2

= 2 (2 - 3)

= \frac{2 ( 2 - 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 - 3

= - (2 - 3)

Distribute parentheses

= - (2) - (- 3)

Apply minus-plus rules

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

= - 2 + 3

= - 2 + 3

= - 2 + 3

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Frequently Asked Questions (FAQ)

What is the value of (tan(120)+tan(45))/(1-tan(120)tan(45)) ?
The value of (tan(120)+tan(45))/(1-tan(120)tan(45)) is -2+sqrt(3)