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

The answer to whether tan(pi/2-u)=cot(u) 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(pi/2-u)=cot(u)

Solution

prove

tan (\frac{π}{2} - u) = cot (u)

Solution

True

Solution steps

tan (\frac{π}{2} - u) = cot (u)

Manipulating left side

tan (\frac{π}{2} - u)

Rewrite using trig identities

tan (\frac{π}{2} - u)

Use the basic trigonometric identity:

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

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

Use the Angle Difference identity:

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

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

Use the Angle Difference identity:

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

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

Simplify

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

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

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

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

sin (\frac{π}{2}) cos (u) = cos (u)

sin (\frac{π}{2}) cos (u)

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 (u)

Multiply:

1 \cdot cos (u) = cos (u)

= cos (u)

cos (\frac{π}{2}) sin (u) = 0

cos (\frac{π}{2}) sin (u)

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 (u)

Apply rule

0 \cdot a = 0

= 0

= cos (u) - 0

cos (u) - 0 = cos (u)

= cos (u)

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

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

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

cos (\frac{π}{2}) cos (u) = 0

cos (\frac{π}{2}) cos (u)

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 (u)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (u) = sin (u)

sin (\frac{π}{2}) sin (u)

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 (u)

Multiply:

1 \cdot sin (u) = sin (u)

= sin (u)

= 0 + sin (u)

0 + sin (u) = sin (u)

= sin (u)

= \frac{cos ( u )}{sin ( u )}

= \frac{cos ( u )}{sin ( u )}

= \frac{cos ( u )}{sin ( u )}

Use the basic trigonometric identity:

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

= cot (u)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove csc(θ)-cot(θ)=(sin(θ))/(1+cos(θ))

prove tan(2pi-θ)=-tan(θ)

prove sin^4(θ)-cos^4(θ)=1-2cos^2(θ)

prove (1+tan(x))/(1-tan(x))+(1+cot(x))/(1-cot(x))=0

prove cos(4x)=1-8sin^2(x)cos^2(x)

Frequently Asked Questions (FAQ)

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