Is tan(x-pi/2)=-cot(x) ?

The answer to whether tan(x-pi/2)=-cot(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

prove

tan (x - \frac{π}{2}) = - cot (x)

Solution

True

Solution steps

tan (x - \frac{π}{2}) = - cot (x)

Manipulating left side

tan (x - \frac{π}{2})

Rewrite using trig identities

tan (x - \frac{π}{2})

Use the basic trigonometric identity:

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

= \frac{sin ( x - \frac{π}{2} )}{cos ( x - \frac{π}{2} )}

Use the Angle Difference identity:

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

= \frac{sin ( x ) cos ( \frac{π}{2} ) - cos ( x ) sin ( \frac{π}{2} )}{cos ( x - \frac{π}{2} )}

Use the Angle Difference identity:

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

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

Simplify

\frac{sin ( x ) cos ( \frac{π}{2} ) - cos ( x ) sin ( \frac{π}{2} )}{cos ( x ) cos ( \frac{π}{2} ) + sin ( x ) sin ( \frac{π}{2} )} : - \frac{cos ( x )}{sin ( x )}

\frac{sin ( x ) cos ( \frac{π}{2} ) - cos ( x ) sin ( \frac{π}{2} )}{cos ( x ) cos ( \frac{π}{2} ) + sin ( x ) sin ( \frac{π}{2} )}

sin (x) cos (\frac{π}{2}) - cos (x) sin (\frac{π}{2}) = - cos (x)

sin (x) cos (\frac{π}{2}) - cos (x) sin (\frac{π}{2})

sin (x) cos (\frac{π}{2}) = 0

sin (x) cos (\frac{π}{2})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

cos (x) sin (\frac{π}{2}) = cos (x)

cos (x) sin (\frac{π}{2})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot cos (x)

Multiply:

cos (x) \cdot 1 = cos (x)

= cos (x)

= 0 - cos (x)

0 - cos (x) = - cos (x)

= - cos (x)

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

cos (x) cos (\frac{π}{2}) + sin (x) sin (\frac{π}{2}) = sin (x)

cos (x) cos (\frac{π}{2}) + sin (x) sin (\frac{π}{2})

cos (x) cos (\frac{π}{2}) = 0

cos (x) cos (\frac{π}{2})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (x) sin (\frac{π}{2}) = sin (x)

sin (x) sin (\frac{π}{2})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot sin (x)

Multiply:

sin (x) \cdot 1 = sin (x)

= sin (x)

= 0 + sin (x)

0 + sin (x) = sin (x)

= sin (x)

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

Apply the fraction rule:

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

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

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

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

Use the basic trigonometric identity:

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

= - cot (x)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove tan^2(x)=(1-cos(2x))/(1+cos(2x))prove

tan^{2} (x) = \frac{1 - cos ( 2 x )}{1 + cos ( 2 x )}

prove sec(θ)-sin(θ)tan(θ)=cos(θ)prove

sec (θ) - sin (θ) tan (θ) = cos (θ)

prove sin(pi-θ)=sin(θ)prove

sin (π - θ) = sin (θ)

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

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

prove tan(x)cot(x)=1 prove

tan (x) cot (x) = 1

Frequently Asked Questions (FAQ)

Is tan(x-pi/2)=-cot(x) ?
The answer to whether tan(x-pi/2)=-cot(x) is True