Is tan(135+θ)=(-1+tan(θ))/(1+tan(θ)) ?

The answer to whether tan(135+θ)=(-1+tan(θ))/(1+tan(θ)) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan(135+θ)=(-1+tan(θ))/(1+tan(θ))

Solution

prove

tan (13 5^{\circ} + θ) = \frac{- 1 + tan ( θ )}{1 + tan ( θ )}

Solution

True

Solution steps

tan (13 5^{\circ} + θ) = \frac{- 1 + tan ( θ )}{1 + tan ( θ )}

Manipulating left side

tan (13 5^{\circ} + θ)

Rewrite using trig identities

tan (13 5^{\circ} + θ)

Use the basic trigonometric identity:

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

= \frac{sin ( 13 5 ^{\circ} + θ )}{cos ( 13 5 ^{\circ} + θ )}

Use the Angle Sum identity:

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

= \frac{sin ( 13 5 ^{\circ} ) cos ( θ ) + cos ( 13 5 ^{\circ} ) sin ( θ )}{cos ( 13 5 ^{\circ} + θ )}

Use the Angle Sum identity:

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

= \frac{sin ( 13 5 ^{\circ} ) cos ( θ ) + cos ( 13 5 ^{\circ} ) sin ( θ )}{cos ( 13 5 ^{\circ} ) cos ( θ ) - sin ( 13 5 ^{\circ} ) sin ( θ )}

Simplify

\frac{sin ( 13 5 ^{\circ} ) cos ( θ ) + cos ( 13 5 ^{\circ} ) sin ( θ )}{cos ( 13 5 ^{\circ} ) cos ( θ ) - sin ( 13 5 ^{\circ} ) sin ( θ )} : - \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

\frac{sin ( 13 5 ^{\circ} ) cos ( θ ) + cos ( 13 5 ^{\circ} ) sin ( θ )}{cos ( 13 5 ^{\circ} ) cos ( θ ) - sin ( 13 5 ^{\circ} ) sin ( θ )}

sin (13 5^{\circ}) cos (θ) + cos (13 5^{\circ}) sin (θ) = \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

sin (13 5^{\circ}) cos (θ) + cos (13 5^{\circ}) sin (θ)

Simplify

sin (13 5^{\circ}) : \frac{2}{2}

sin (13 5^{\circ})

Use the following trivial identity:

sin (13 5^{\circ}) = \frac{2}{2}

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{2}{2}

= \frac{2}{2} cos (θ) + cos (13 5^{\circ}) sin (θ)

Simplify

cos (13 5^{\circ}) : - \frac{2}{2}

cos (13 5^{\circ})

Use the following trivial identity:

cos (13 5^{\circ}) = - \frac{2}{2}

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{2}{2}

= \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{cos ( 13 5 ^{\circ} ) cos ( θ ) - sin ( 13 5 ^{\circ} ) sin ( θ )}

cos (13 5^{\circ}) cos (θ) - sin (13 5^{\circ}) sin (θ) = - \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

cos (13 5^{\circ}) cos (θ) - sin (13 5^{\circ}) sin (θ)

Simplify

cos (13 5^{\circ}) : - \frac{2}{2}

cos (13 5^{\circ})

Use the following trivial identity:

cos (13 5^{\circ}) = - \frac{2}{2}

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{2}{2}

= - \frac{2}{2} cos (θ) - sin (13 5^{\circ}) sin (θ)

Simplify

sin (13 5^{\circ}) : \frac{2}{2}

sin (13 5^{\circ})

Use the following trivial identity:

sin (13 5^{\circ}) = \frac{2}{2}

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{2}{2}

= - \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{- \frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}

Multiply

\frac{2}{2} cos (θ) : \frac{2 cos ( θ )}{2}

\frac{2}{2} cos (θ)

Multiply fractions:

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

= \frac{2 cos ( θ )}{2}

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{- \frac{2 c o s ( θ )}{2} - \frac{2}{2} sin ( θ )}

Multiply

\frac{2}{2} sin (θ) : \frac{2 sin ( θ )}{2}

\frac{2}{2} sin (θ)

Multiply fractions:

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

= \frac{2 sin ( θ )}{2}

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{- \frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}

Multiply

\frac{2}{2} cos (θ) : \frac{2 cos ( θ )}{2}

\frac{2}{2} cos (θ)

Multiply fractions:

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

= \frac{2 cos ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2}{2} sin ( θ )}{- \frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}

Multiply

\frac{2}{2} sin (θ) : \frac{2 sin ( θ )}{2}

\frac{2}{2} sin (θ)

Multiply fractions:

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

= \frac{2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}{- \frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}

Combine the fractions

- \frac{2 cos ( θ )}{2} - \frac{2 sin ( θ )}{2} : \frac{- 2 cos ( θ ) - 2 sin ( θ )}{2}

Apply rule

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

= \frac{- 2 cos ( θ ) - 2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}{\frac{- 2 c o s ( θ ) - 2 s i n ( θ )}{2}}

Combine the fractions

\frac{2 cos ( θ )}{2} - \frac{2 sin ( θ )}{2} : \frac{2 cos ( θ ) - 2 sin ( θ )}{2}

Apply rule

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

= \frac{2 cos ( θ ) - 2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ ) - 2 s i n ( θ )}{2}}{\frac{- 2 c o s ( θ ) - 2 s i n ( θ )}{2}}

Divide fractions:

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

= \frac{( 2 cos ( θ ) - 2 sin ( θ ) ) \cdot 2}{2 ( - 2 cos ( θ ) - 2 sin ( θ ) )}

Cancel the common factor:

2

= \frac{2 cos ( θ ) - 2 sin ( θ )}{- 2 cos ( θ ) - 2 sin ( θ )}

Factor out common term

2

= \frac{2 ( cos ( θ ) - sin ( θ ) )}{- 2 cos ( θ ) - 2 sin ( θ )}

Factor out common term

2

= - \frac{2 ( cos ( θ ) - sin ( θ ) )}{2 ( cos ( θ ) + sin ( θ ) )}

Cancel the common factor:

2

= - \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

= - \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

= - \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

= \frac{- ( cos ( θ ) - sin ( θ ) )}{cos ( θ ) + sin ( θ )}

Simplify

= \frac{- cos ( θ ) + sin ( θ )}{cos ( θ ) + sin ( θ )}

Manipulating right side

\frac{- 1 + tan ( θ )}{1 + tan ( θ )}

Express with sin, cos

\frac{- 1 + tan ( θ )}{1 + tan ( θ )}

Use the basic trigonometric identity:

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

= \frac{- 1 + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}}

Simplify

\frac{- 1 + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}} : \frac{- cos ( θ ) + sin ( θ )}{cos ( θ ) + sin ( θ )}

\frac{- 1 + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}}

Join

1 + \frac{sin ( θ )}{cos ( θ )} : \frac{cos ( θ ) + sin ( θ )}{cos ( θ )}

1 + \frac{sin ( θ )}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

= \frac{1 \cdot cos ( θ )}{cos ( θ )} + \frac{sin ( θ )}{cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot cos ( θ ) + sin ( θ )}{cos ( θ )}

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + sin ( θ )}{cos ( θ )}

= \frac{- 1 + \frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}

Join

- 1 + \frac{sin ( θ )}{cos ( θ )} : \frac{- cos ( θ ) + sin ( θ )}{cos ( θ )}

- 1 + \frac{sin ( θ )}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

= - \frac{1 \cdot cos ( θ )}{cos ( θ )} + \frac{sin ( θ )}{cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot cos ( θ ) + sin ( θ )}{cos ( θ )}

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{- cos ( θ ) + sin ( θ )}{cos ( θ )}

= \frac{\frac{- c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}

Divide fractions:

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

= \frac{( - cos ( θ ) + sin ( θ ) ) cos ( θ )}{cos ( θ ) ( cos ( θ ) + sin ( θ ) )}

Cancel the common factor:

cos (θ)

= \frac{- cos ( θ ) + sin ( θ )}{cos ( θ ) + sin ( θ )}

= \frac{- cos ( θ ) + sin ( θ )}{cos ( θ ) + sin ( θ )}

= \frac{- cos ( θ ) + sin ( θ )}{cos ( θ ) + sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove (csc(pi/2-x))/(-tan(-x))=csc(x)prove

\frac{csc ( \frac{π}{2} - x )}{- tan ( - x )} = csc (x)

prove cot^2(θ)cos^2(θ)=cot^2(θ)-cos^2(θ)prove

cot^{2} (θ) cos^{2} (θ) = cot^{2} (θ) - cos^{2} (θ)

prove tan(120)=tan(180-60)prove

tan (12 0^{\circ}) = tan (18 0^{\circ} - 6 0^{\circ})

prove sin(2x)tan(x)=2sin^2(x)prove

sin (2 x) tan (x) = 2 sin^{2} (x)

prove cot(b)sec(b)sin(b)=1 prove

cot (b) sec (b) sin (b) = 1

Frequently Asked Questions (FAQ)

Is tan(135+θ)=(-1+tan(θ))/(1+tan(θ)) ?
The answer to whether tan(135+θ)=(-1+tan(θ))/(1+tan(θ)) is True