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 Trigonometry >

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 Sum 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 Sum 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 (cos(pi-x))/(cos(pi/2+x))=cot(x)

prove cot(θ)+tan(θ)=2csc(2θ)

prove cos(x)(tan(x)-sec(-x))=sin(x)-1

prove 1/(sec^2(x))+1/(csc^2(x))=1

prove (1+tan^2(x))cot^2(x)=csc^2(x)

Frequently Asked Questions (FAQ)

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