Is tan((5pi)/(12))=tan(pi/4+pi/6) ?

The answer to whether tan((5pi)/(12))=tan(pi/4+pi/6) is True

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prove tan((5pi)/(12))=tan(pi/4+pi/6)

Solution

prove

tan (\frac{5 π}{12}) = tan (\frac{π}{4} + \frac{π}{6})

Solution

True

Solution steps

tan (\frac{5 π}{12}) = tan (\frac{π}{4} + \frac{π}{6})

Manipulating left side

tan (\frac{5 π}{12})

Simplify

tan (\frac{5 π}{12}) : 2 + 3

tan (\frac{5 π}{12})

Rewrite using trig identities:

\frac{tan ( \frac{π}{4} ) + tan ( \frac{π}{6} )}{1 - tan ( \frac{π}{4} ) tan ( \frac{π}{6} )}

tan (\frac{5 π}{12})

Write

tan (\frac{5 π}{12})

tan (\frac{π}{4} + \frac{π}{6})

= tan (\frac{π}{4} + \frac{π}{6})

Use the Angle Sum identity:

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

= \frac{tan ( \frac{π}{4} ) + tan ( \frac{π}{6} )}{1 - tan ( \frac{π}{4} ) tan ( \frac{π}{6} )}

= \frac{tan ( \frac{π}{4} ) + tan ( \frac{π}{6} )}{1 - tan ( \frac{π}{4} ) tan ( \frac{π}{6} )}

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

Use the following trivial identity:

tan (\frac{π}{6}) = \frac{3}{3}

tan (\frac{π}{6})

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= \frac{3}{3}

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

Simplify

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

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

Multiply:

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

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

Join

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

1 - \frac{3}{3}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3 - 3}{3}

Factor

3 - 3 : 3 (3 - 1)

3 - 3

3 = 33

= 33 - 3

Factor out common term

3

= 3 (3 - 1)

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

Cancel

\frac{3 ( 3 - 1 )}{3} : \frac{3 - 1}{3}

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

Apply radical rule:

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

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= \frac{3 - 1}{3}

= \frac{3 - 1}{3}

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

Join

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

1 + \frac{3}{3}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3 + 3}{3}

Factor

3 + 3 : 3 (3 + 1)

3 + 3

3 = 33

= 33 + 3

Factor out common term

3

= 3 (3 + 1)

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

Cancel

\frac{3 ( 3 + 1 )}{3} : \frac{3 + 1}{3}

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

Apply radical rule:

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

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= \frac{3 + 1}{3}

= \frac{3 + 1}{3}

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

Divide fractions:

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

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

Cancel the common factor:

3

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

Rationalize

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

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

Multiply by the conjugate

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

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

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

(3 + 1) (3 + 1)

Apply exponent rule:

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

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

(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

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

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

(3 - 1) (3 + 1)

Apply Difference of Two Squares Formula:

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

a = 3, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (3)^{2} - 1

(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

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= 2

= \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

= 2 + 3

= 2 + 3

Manipulating right side

tan (\frac{π}{4} + \frac{π}{6})

Rewrite using trig identities

tan (\frac{π}{4} + \frac{π}{6})

Use the basic trigonometric identity:

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

= \frac{sin ( \frac{π}{4} + \frac{π}{6} )}{cos ( \frac{π}{4} + \frac{π}{6} )}

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= \frac{sin ( \frac{π}{4} ) cos ( \frac{π}{6} ) + cos ( \frac{π}{4} ) sin ( \frac{π}{6} )}{cos ( \frac{π}{4} + \frac{π}{6} )}

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= \frac{sin ( \frac{π}{4} ) cos ( \frac{π}{6} ) + cos ( \frac{π}{4} ) sin ( \frac{π}{6} )}{cos ( \frac{π}{4} ) cos ( \frac{π}{6} ) - sin ( \frac{π}{4} ) sin ( \frac{π}{6} )}

\frac{sin ( \frac{π}{4} ) cos ( \frac{π}{6} ) + cos ( \frac{π}{4} ) sin ( \frac{π}{6} )}{cos ( \frac{π}{4} ) cos ( \frac{π}{6} ) - sin ( \frac{π}{4} ) sin ( \frac{π}{6} )} = 2 + 3

\frac{sin ( \frac{π}{4} ) cos ( \frac{π}{6} ) + cos ( \frac{π}{4} ) sin ( \frac{π}{6} )}{cos ( \frac{π}{4} ) cos ( \frac{π}{6} ) - sin ( \frac{π}{4} ) sin ( \frac{π}{6} )}

sin (\frac{π}{4}) cos (\frac{π}{6}) + cos (\frac{π}{4}) sin (\frac{π}{6}) = \frac{2}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{2}{2}

sin (\frac{π}{4}) cos (\frac{π}{6}) + cos (\frac{π}{4}) sin (\frac{π}{6})

sin (\frac{π}{4}) cos (\frac{π}{6}) = \frac{2}{2} \cdot \frac{3}{2}

sin (\frac{π}{4}) cos (\frac{π}{6})

Simplify

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} cos (\frac{π}{6})

Simplify

cos (\frac{π}{6}) : \frac{3}{2}

cos (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

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

cos (\frac{π}{4}) sin (\frac{π}{6}) = \frac{1}{2} \cdot \frac{2}{2}

cos (\frac{π}{4}) sin (\frac{π}{6})

Simplify

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} sin (\frac{π}{6})

Simplify

sin (\frac{π}{6}) : \frac{1}{2}

sin (\frac{π}{6})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

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

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

= \frac{\frac{2}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{2}{2}}{cos ( \frac{π}{4} ) cos ( \frac{π}{6} ) - sin ( \frac{π}{4} ) sin ( \frac{π}{6} )}

cos (\frac{π}{4}) cos (\frac{π}{6}) - sin (\frac{π}{4}) sin (\frac{π}{6}) = \frac{2}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{2}{2}

cos (\frac{π}{4}) cos (\frac{π}{6}) - sin (\frac{π}{4}) sin (\frac{π}{6})

cos (\frac{π}{4}) cos (\frac{π}{6}) = \frac{2}{2} \cdot \frac{3}{2}

cos (\frac{π}{4}) cos (\frac{π}{6})

Simplify

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} cos (\frac{π}{6})

Simplify

cos (\frac{π}{6}) : \frac{3}{2}

cos (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

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

sin (\frac{π}{4}) sin (\frac{π}{6}) = \frac{1}{2} \cdot \frac{2}{2}

sin (\frac{π}{4}) sin (\frac{π}{6})

Simplify

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} sin (\frac{π}{6})

Simplify

sin (\frac{π}{6}) : \frac{1}{2}

sin (\frac{π}{6})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

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

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

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

Simplify

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

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

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

Multiply fractions:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Simplify

23 : 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

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

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

Multiply fractions:

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

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

Multiply:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{\frac{2}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{2}{2}}{\frac{6}{4} - \frac{2}{4}}

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

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

Multiply fractions:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Simplify

23 : 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

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

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

Multiply fractions:

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

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

Multiply:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{\frac{6}{4} + \frac{2}{4}}{\frac{6}{4} - \frac{2}{4}}

Combine the fractions

\frac{6}{4} - \frac{2}{4} : \frac{6 - 2}{4}

Apply rule

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

= \frac{6 - 2}{4}

= \frac{\frac{6}{4} + \frac{2}{4}}{\frac{6 - 2}{4}}

Combine the fractions

\frac{6}{4} + \frac{2}{4} : \frac{6 + 2}{4}

Apply rule

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

= \frac{6 + 2}{4}

= \frac{\frac{6 + 2}{4}}{\frac{6 - 2}{4}}

Divide fractions:

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

= \frac{( 6 + 2 ) \cdot 4}{4 ( 6 - 2 )}

Cancel the common factor:

4

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

Rationalize

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

\frac{6 + 2}{6 - 2}

Multiply by the conjugate

\frac{6 + 2}{6 + 2}

= \frac{( 6 + 2 ) ( 6 + 2 )}{( 6 - 2 ) ( 6 + 2 )}

(6 + 2) (6 + 2) = 8 + 43

(6 + 2) (6 + 2)

Apply exponent rule:

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

(6 + 2) (6 + 2) = (6 + 2)^{1 + 1}

= (6 + 2)^{1 + 1}

Add the numbers:

1 + 1 = 2

= (6 + 2)^{2}

Apply Perfect Square Formula:

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

a = 6, b = 2

= (6)^{2} + 262 + (2)^{2}

Simplify

(6)^{2} + 262 + (2)^{2} : 8 + 43

(6)^{2} + 262 + (2)^{2}

(6)^{2} = 6

(6)^{2}

Apply radical rule:

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

= (6^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 6^{\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

= 6

262 = 43

262

Factor integer

6 = 2 \cdot 3

= 2 2 \cdot 3 2

Apply radical rule:

2 \cdot 3 = 23

= 2232

Apply radical rule:

a a = a

22 = 2

= 2 \cdot 23

Multiply the numbers:

2 \cdot 2 = 4

= 43

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

= 6 + 43 + 2

Add the numbers:

6 + 2 = 8

= 8 + 43

= 8 + 43

(6 - 2) (6 + 2) = 4

(6 - 2) (6 + 2)

Apply Difference of Two Squares Formula:

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

a = 6, b = 2

= (6)^{2} - (2)^{2}

Simplify

(6)^{2} - (2)^{2} : 4

(6)^{2} - (2)^{2}

(6)^{2} = 6

(6)^{2}

Apply radical rule:

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

= (6^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 6^{\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

= 6

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

= 6 - 2

Subtract the numbers:

6 - 2 = 4

= 4

= 4

= \frac{8 + 4 3}{4}

Factor

8 + 43 : 4 (2 + 3)

8 + 43

Rewrite as

= 4 \cdot 2 + 43

Factor out common term

4

= 4 (2 + 3)

= \frac{4 ( 2 + 3 )}{4}

Divide the numbers:

\frac{4}{4} = 1

= 2 + 3

= 2 + 3

= 2 + 3

= 2 + 3

= 2 + 3

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove sin^2(θ)=csc^2(θ)prove

sin^{2} (θ) = csc^{2} (θ)

prove tan(x)(sec^2(x)-1)=tan^3(x)prove

tan (x) (sec^{2} (x) - 1) = tan^{3} (x)

prove sin(x)=-1 prove

sin (x) = - 1

prove sec(b)+tan(b)=(cos(b))/(1-sin(x))prove

sec (b) + tan (b) = \frac{cos ( b )}{1 - sin ( x )}

prove 2sinh(ln(x))+(1^2)/2-4/3 =2 prove

2 sinh (ln (x)) + \frac{1 ^{2}}{2} - \frac{4}{3} = 2

Frequently Asked Questions (FAQ)

Is tan((5pi)/(12))=tan(pi/4+pi/6) ?
The answer to whether tan((5pi)/(12))=tan(pi/4+pi/6) is True