Cotangent Formula

Trigonometry is an important branch of mathematics that deals with the relation between the lengths of sides and angles of a right-angled triangle. Sine, Cosine, tangent, cosecant, secant, and cotangent are the six trigonometric ratios or functions. Where a trigonometric ratio is depicted as the ratio between the sides of a right-angled triangle. 

• sin θ = opposite side/hypotenuse
• cos θ = adjacent side/hypotenuse
• tan θ = opposite side/adjacent side
• cosec θ = 1/sin θ = hypotenuse/opposite side
• sec θ = 1/cos θ = hypotenuse/adjacent side
• cot θ = 1/tan θ = adjacent side/opposite side

Cotangent Formula

A Cotangent function is a reciprocal function of the given tangent function. The value of a cotangent angle in a right-angled triangle is the ratio of the length of the side adjacent to the given angle to the length of the side opposite to the given angle. We write cotangent function as “cot”.

Triangle ABC

Now, the cotangent formula for the angle θ is,

cot θ = (Adjacent side)/(Opposite side)

• The cotangent function is positive in the first and third quadrants and negative in the second and fourth quadrants.

1. cot (2π + θ) = cot θ (1^st quadrant)
2. cot (π – θ) = – cot θ (2^nd quadrant)
3. cot (π + θ) = cot θ (3^rd quadrant)
4. cot (2π – θ) = – cot θ (4^th quadrant)

• The cotangent function is a negative function since the cotangent of a negative angle is the negative of a cotangent positive angle.

cot (-θ) = – cot θ

• In terms of the tangent function, the cotangent function is written as,

cot θ = 1/tan θ

(or)

cot θ = tan (90° – θ) (or) tan (π/2 – θ)

• The cotangent function in terms of sine and cosine functions can be written as,

cot θ = cos θ/sin θ

We know that, cot θ = adjacent side/opposite side

Now divide both the numerator and denominator with hypotenuse

⇒ cot θ = (adjacent side/hypotenuse) / (opposite side/hypotenuse)

We know that, sin θ = opposite side/hypotenuse

cos θ = adjacent side/hypotenuse

Hence, cot θ = cos θ/sin θ

• Cotangent function in terms of sine function can be written as,

cot θ = (√1 – sin² θ)/sin θ 

We know that, cot θ = cos θ/sin θ

From the Pythagorean identities we have;

cos² θ + sin² θ = 1

⇒ cos θ = √1 – sin² θ

Hence, cot θ =

• Cotangent function in terms of cosine function can be written as,

cot θ = cos θ/(√1 -cos² θ)

We know that, cot θ = cos θ/sin θ

From the Pythagorean identities we have;

cos² θ + sin² θ = 1

sin θ = √1 – cos² θ

Hence, cot θ = 

• Cotangent function in terms of secant and cosecant functions can be written as,

cot θ = cosec θ/sec θ

We have, cot θ = cos θ/sin θ

This can be written as, cot θ = (1/sin θ) / (1/cos θ)

⇒ cot θ = cosec θ/sec θ

• Cotangent function in terms of cosecant function can be written as:

cot θ = √(cosec² – 1)

From the Pythagorean identities, we have,

cosec² θ – cot² θ = 1

⇒ cot² θ = 1 – cosec² – 1

Hence, cot θ = √(cosec² – 1)

• Cotangent function in terms of secant function can be written as:

cot θ = 1/(√sec² θ – 1)

From the Pythagorean identities, we have,

sec² θ – tan² θ = 1

tan θ = √sec² θ – 1

We know that, cot θ = 1/tan θ

Hence, cot θ =

Trigonometric Ratio table

Trigonometric Ratio Table

Cotangent law or Law of Cotangents

Cotangent law looks similar to sine law, but here it involves half angles. The law of cotangents describes the relationship between the lengths of the sides of the triangle and the cotangents of the halves of the three angles. Consider a triangle ABC, where a, b, and c are the lengths of the sides of the triangle. 

The law of cotangents states that,

Where s is the semi-perimeter of the triangle ABC and r is its inradius of the inscribed circle of the triangle.

s = (a + b + c)/2

r =

Sample Problems

Problem 1: Find the value of cot θ if tan θ = 3/4.

Solution:

Given data, tan θ = 3/4

We know that, cot θ = 1/tan θ

⇒ cot θ = 1/(3/4) = 4/3

So, cot θ = 4/3

Problem 2: Find the value of cot α, sin α = 1/3, and cos α = 2√2/3.

Solution:

Given data, sin α = 1/3 and cos α = 2√2/3

We know that, cot α = cos α/sin α

⇒ cot α = (2√2/3) / (1/3) = 2√2

Hence, the value of cot α = 2√2

Problem 3: A boy standing 15 m from a tree is looking at a 30-degree angle to the top of the tree. What is the height of the tree?

Solution:

Diagram from the given data

Given data, the distance between the boy and the foot of the tree = 15 m and θ = 30°

Let the height of the tree be ‘h’

We have, cot θ = adjacent side/opposite side

⇒ cot 30° = 15/h

⇒ √3 = 15/h [since, cot 30° = √3] 

⇒ h = 15/√3 

⇒ h = 5√3 m

Hence, the height of the tree = 5√3 m

Problem 4: Find the value of cot x if sec x = 6/5.

Solution:

Given data, sec x = 6/5

We have, sec² x – tan² x = 1

⇒ (6/5)² – tan² x = 1

⇒ 36/25 – tan² x = 1

⇒ tan² x = 36/25 – 1

⇒ tan² x = 11/25 

⇒ tan x = √(11/25) = √11/5

We know that, cot x = 1/tan x

⇒ cot x = 1/(√11/5) = 5/√11

Hence, cot x = 5/√11

Problem 5: Find the value of cot θ if cosec θ = 25/24.

Solution:

Given data, cosec θ = 25/24

We know that, cot θ = √(cosec² – 1)

⇒ cot θ = √(25/24)² – 1

⇒ cot θ =√(625 – 576)/576 = √49/576

⇒ cot θ = 7/24

Hence, the value of cot θ = 7/24

Problem 6: Find the value of cot β if sin β = 5/13.

Solution:

Given data, sin β = 5/13

We know that, sin² β + cos² β = 1

⇒ (5/13)² + cos² β = 1

⇒ cos² β = 1 – (5/13)² = 1 – 25/169 = 144/169

⇒ cos β = √144/169 = 12/13

cot β = cosβ/sin β

= (12/13) / (5/13)

⇒ cot β = 12/5

Hence, the value of cot β = 12/5

Problem 7: Using the law of cotangents, find the values of ∠A, ∠B, and ∠C (in degrees) if the lengths of the three sides of triangle ABC are a = 4 cm, b= 3 cm, and c= 3 cm.

Solution: 

Given, a = 4 cm, b = 3 cm and c = 3 cm

Triangle ABC

From the cotangents law, 

s = (a + b + c)/2

⇒ s = (3 + 4 + 3)/2 = 10/2 = 5

Now, s – a = 5 – 4 = 1

⇒ s – b = 5 – 3 = 2

⇒ s – c = 5 – 3 = 2

r =

⇒ r = √[(1)(2)(2)/5]

Inradius of the triangle r = 2/√5

From the equation of law of cotangents,

cot (A/2)/1 = 1/(2/√5)

⇒ cot (A/2) = √5/2 ⇒ A/2 = cot ^-1 (√5/2)

⇒ (A/2) = 41.8° ⇒ ∠A = 83.6°

cot(B/2)/2 = 1/(2/√5)

⇒ cot(B/2)/2 = √5/2 ⇒ cot (B/2) = √5

⇒ (B/2) = cot^-1 (√5) = 24.1° ⇒ ∠B = 48.2°

cot (C/2)/2 =  1/(2/√5)

⇒ cot(C/2) = √5 ⇒ (C/2) = cot^-1 (√5)

⇒ (C/2) = 24.1° ⇒ ∠C = 48.2°

Hence, the angles of triangle ABC are  ∠A = 83.6°, ∠B = 48.2° and ∠C = 48.2°.