The tangent formulas are formulas about the tangent function in trigonometry. The tangent function (which is usually referred to as “tan”) is one of the 6 trigonometric functions which is the ratio of the opposite side to the adjacent side. There are multiple formulas related to tangent function which can be derived from various trigonometric identities and formulas. Let us learn the tangent formulas along with a few solved examples.

## What Are Tangent Formulas?

The tangent formulas talk about the tangent (tan) function. Let us consider a right-angled triangle with one of its acute angles to be x. Then the tangent formula is, tan x = (opposite side) / (adjacent side), where “opposite side” is the side opposite to the angle x, and “adjacent side” is the side that is adjacent to the angle x. Apart from this general formula, there are so many other formulas in trigonometry that will define a tangent function which you can see in the following image.

### Tangent Formulas Using Reciprocal Identity

We know that the tangent function (tan) and the cotangent function (cot) are reciprocals of each other. i.e., if tan x = a / b, then cot x = b / a. Thus, tangent formula using one of the reciprocal identities is,

**tan x = 1 / (cot x)**

### Tangent Formula Using Sin and Cos

We know that sin x = (opposite) / (hypotenuse), cos x = (adjacent) / (hypotenuse), and tan x = (opposite) / (adjacent). Now we will divide sin x by cos x.

(sin x) / (cos x) = [ (opposite) / (hypotenuse) ] / [ (adjacent) / (hypotenuse) ] = (opposite) / (adjacent) = tan x

Thus, the tangent formula in terms of sine and cosine is,

**tan x = (sin x) / (cos x)**

### Tangent Formulas Using Pythagorean Identity

One of the Pythagorean identities talks about the relationship between sec and tan. It says, sec^{2}x – tan^{2}x = 1, for any x. We can solve this for tan x. Let us see how.

sec^{2}x – tan^{2}x = 1

Subtracting sec^{2}x from both sides,

-tan^{2}x = 1 – sec^{2}x

Multiplying both sides by -1,

tan^{2}x = sec^{2}x – 1

Taking square root on both sides,

**tan x = ± √( sec ^{2}x – 1)**

### Tangent Formula Using Cofunction Identities

The cofunction identities define the relation between the cofunctions which are sin, cos; sec, csc; and tan, cot. Using one of the cofunction identities,

- tan x = cot (90
^{o}– x) (OR) - tan x = cot (π/2 – x)

### Tangent Formulas Using Sum/Difference Formulas

We have sum/difference formulas for every trigonometric function that deal with the sum of angles (A + B) and the difference of angles (A – B). The sum/difference formulas of tangent function are,

- tan (A + B) = (tan A + tan B) / (1 – tan A tan B)
- tan (A – B) = (tan A – tan B) / (1 + tan A tan B)

### Tangent Formula of Double Angle

We have double angle formulas in trigonometry which deal with 2 times the angle. The double angle formula of tan is

tan 2x = (2 tan x) / (1 – tan^{2}x)

### Tangent Formula of Triple Angle

We have triple angle formulas for all trigonometric functions. Among them, the triple angle formula of the tangent function is,

tan 3x = (3 tan x – tan^{3}x) / (1 – 3 tan^{2}x)

### Tangent Formula of Half Angle

We have half-angle formulas in trigonometry that deal with half of the angles (x/2). The half-angle formulas of tangent function are,

- tan (x/2) =± √[ (1 – cos x) / (1 + cos x) ]
- tan (x/2) = (1 – cos x) / ( sin x)

## Examples Using Tangent Formulas

**Example 1: **If sec x = 5/3 and x is in the first quadrant, find the value of tan x.

**Solution:**

Using one of the tangent formulas,

tan x = ± √(sec^{2}x – 1)

Since x is in the first quadrant, cos x is positive. Thus,

tan x = ± √(sec^{2}x – 1)

Substitute sec x = 5/3 here,

tan x = √((5/3)^{2} – 1)

= √((25/9) – 1)

=√ (16/9)

= 4/3

**Answer: **tan x = 4/3.

**Example 2: **If cot (90 – A) = 3/2, then find the value of tan A.

**Solution:**

Using one of the tangent formulas,

tan A = cot (90 – A)

It is given that cot (90 – A) = 3/2. Hence,

tan A = 3/2

**Answer: **tan A = 3/2.

**Example 3: **If tan A = 1/2 and tan B = 1/3, find tan (A + B).

**Solution:**

Using the tangent formula of addition,

tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

= (1/2 + 1/3) / ( 1 – (1/2) · (1/3) )

= (5/6) / (1 – (1/6))

= (5/6) / (5/6)

= 1

**Answer: **tan (A + B) = 1.

## FAQs on Tangent Formulas

### What Are Tangent Formulas?

The tangent formulas are related to the tangent function. The important tangent formulas are as follows:

- tan x = (opposite side) / (adjacent side)
- tan x = 1 / (cot x)
- tan x = (sin x) / (cos x)
- tan x = ± √( sec
^{2}x – 1)

### How To Derive Tangent Formula of Sum?

The tangent formula of sum/addition is, tan (A + B) = (tan A + tan B) / (1 – tan A tan B). Let us derive this starting with the left side part.

tan (A + B) = sin (A + B) / cos (A + B)

= [ sin A cos B + cos A sin B ] / [cos A cos B – sin A sin B]

Dividing each term in both numerator and denominator by cos A cos B,

tan (A + B) = [ (sin A / cos A) + (sin B / cos B) ] / [ 1 – (sin A / cos A) (sin B / cos B) ]

= (tan A + tan B) / (1 – tan A tan B)

### What Are the Applications of Tangent Formulas?

As we have learned on this page, we have multiple tangent formulas and we can choose one of them to prove a trigonometric identity (or) find the value of the tangent function with the available information. We also use tangent formulas in Calculus.

### How To Derive the Double Angle Tangent Formula?

Using the sum formula of tangent function, we have, tan (A + B) = (tan A + tan B) / (1 – tan A tan B). Substituting A = B on both sides here, we get,

tan 2x = (tan x + tan x) / (1 – tan x · tan x) = (2 tan x) / (1 – tan^{2}x).

**Some Basic Tangent Formulae**

**Tangent Function in Quadrants**

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

tan (2π + θ) = tan θ(1^{st }quadrant)

tan (π – θ) = – tan θ(2^{nd}quadrant)

tan (π + θ) = tan θ(3^{rd}quadrant)

tan (2π – θ) = – tan θ(4^{th}quadrant)

**Tangent Function as a Negative Function**

The tangent function is a negative function since the tangent of a negative angle is the negative of a tangent positive angle.

tan (-θ) = – tan θ

**Tangent Function in Terms of Sine and Cosine Function**

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

tan θ = sin θ/cos θWe know that, tan θ = Opposite side/Adjacent side

Now, divide both the numerator and denominator with hypotenuse

tan θ = (Opposite side/Hypotenuse)/(Adjacent side/Hypotenuse)

We know that, sin θ = opposite side/hypotenuse

cos θ = adjacent side/hypotenuse

Hence, tan θ = sin θ/cos θ

**Tangent Function in Terms of the Sine Function**

The tangent function in terms of the sine function can be written as,

tan θ = sin θ/(√1 – sin^{2 }θ)We know that,

tan θ = sin θ/cos θ

From the Pythagorean identities, we have,

sin

^{2}θ + cos^{2}θ = 1cos

^{2}θ = 1 – sin^{2}θcos θ = √(1 – sin

^{2}θ)Hence, tan θ = sin θ/(√1 – sin

^{2}θ)

**Tangent Function in Terms of the Cosine Function**

The tangent function in terms of the cosine function can be written as,

tan θ = (√1 -cos^{2}θ)/cos θWe know that,

tan θ = sin θ/cos θ

From the Pythagorean identities, we have,

sin

^{2}θ + cos^{2}θ = 1sin

^{2}θ = 1 – cos^{2}θsin θ = √(1 – cos

^{2}θ)Hence, tan θ = (√1 – cos

^{2}θ)/cos θ

**Tangent Function in Terms of the Cotangent Function**

The tangent function in terms of the cotangent function can be written as,

tan θ = 1/cot θor

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

**Tangent Function in Terms of the Cosecant Function**

The tangent function in terms of the cosecant function can be written as,

tan θ = 1/√(cosec^{2 }θ – 1)From the Pythagorean identities, we have,

cosec

^{2}θ – cot^{2}θ = 1cot

^{2}θ = cosec^{2 }θ – 1cot θ = √(cosec

^{2}θ – 1)We know that,

tan θ = 1/cot θ

Hence, tan θ = 1/√(cosec

^{2}θ – 1)

**Tangent Function in Terms of the Secant Function**

The tangent function in terms of the secant function can be written as,

tan θ = √sec^{2}θ – 1From the Pythagorean identities, we have,

sec

^{2}θ – tan^{2}θ = 1tan θ = sec

^{2}θ – 1Hence, tan θ = √(sec

^{2}θ – 1)

**Tangent Function in Terms of Double Angle**

The tangent function for a double angle is,

tan 2θ = (2 tan θ)/(1 – tan^{2}θ)

**Tangent Function in Terms of Triple Angle**

The tangent function for a triple angle is,

tan 3θ = (3 tan θ – tan^{3}θ) / (1 – 3 tan^{2}θ)

**Tangent Function in Terms of Half-Angle**

The tangent function for a half-angle is,

tan (θ/2) = ± √[ (1 – cos θ) / (1 + cos θ) ]

tan (θ/2) = (1 – cos θ) / ( sin θ)

**Tangent Function in Terms of the Addition and Subtraction of Two Angles**

The sum and difference formulas for a tangent function are,

tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

tan (A – B) = (tan A – tan B)/(1 + tan A tan B)

**Trigonometric Ratio Table**

Angle
| Angle
| sin θ | cos θ | tan θ = sin θ/cos θ | cosec θ | sec θ | cot θ |

0° | 0 | 0 | 1 | 0/1 = 0 | Undefined | 1 | Undefined |

30° | π/6 | 1/2 | √3/2 | (1/2)/(√3/2) = 1/√3 | 2 | 2/√3 | √3 |

45° | π/4 | 1/√2 | 1/√2 | (1/√2)/(1/√2) = 1 | √2 | √2 | 1 |

60° | π/3 | √3/2 | 1/2 | (√3/2)/(1/2) = √3 | 2/√3 | 2 | 1/√3 |

90° | π/2 | 1 | 0 | 1/0 = undefined | 1 | Undefined | 0 |

120° | 2π/3 | √3/2 | -1/2 | (√3/2)/(-1/2) = -√3 | 2/√3 | -2 | -1/√3 |

150° | 5π/6 | 1/2 | -(√3/2) | (1/2)/(-√3/2) = -1/√3 | 2 | -(2/√3) | -√3 |

180° | π | 0 | -1 | 0/(-1) = 0 | Undefined | -1 | Undefined |

## Solved Example on Tangent Formulas

**Example 1: Find the value of tan θ if sin θ = 2/5 and θ is the first quadrant angle.**

**Solution:**

Given, sin θ = 2/5

From the Pythagorean identities we have,

sin

^{2}θ + cos^{2}θ = 1cos

^{2}θ = 1 – sin^{2}θ = 1 – (2/5)^{2}cos

^{2}θ = 1 – (4/5) = 21/25cos θ = ±√21/5

Since θ is the first quadrant angle, cos θ is positive.

cos θ = √21/5

We know that,

tan θ = sin θ/cos θ

= (2/3)/(√21/5)

Hence,

tan θ = 10/3√21

**Example 2: Find the value of tan x if sec x = 13/12 and x is the fourth quadrant angle.**

**Solution:**

Given, sec x = 13/12

From the Pythagorean identities, we have,

sec

^{2}x – tan^{2}x = 1tan

^{2}x = sec^{2}x – 1= (13/12)^{2 }– 1tan

^{2}x = (169/144) – 1= 25/144tan x = ± 5/12

Since x is the fourth quadrant angle, tan x is negative.

tan x = – 5/12

Hence,

tan x = – 5/12

**Example 3: If tan X = 2/3 and tan Y = 1/2, then what is the value of tan (X + Y)?**

**Solution:**

Given,

tan X = 2/3 and tan Y = 1/2

We know that,

tan (X + Y) = (tan X + tan Y)/(1 – tan X tan Y)

tan (X + Y) = [(2/3) + (1/2)]/[1 – (2/3)×(1/2)]

= (7/6)/(2/3) = 7/4

Hence,

tan (X + Y) = 7/4

**Example 4: Calculate the tangent function if the adjacent and opposite sides of a right-angled triangle are 4 cm and 7 cm, respectively.**

**Solution:**

Given,

Adjacent side = 4 cm

Opposite side = 7 cm

We know that,

tan θ = Opposite side/Adjacent side

tan θ = 7/4 = 1.75

Hence,

tan θ = 1.75

**Example 5: A man is looking at a clock tower at a 60° angle to the top of the tower, whose height is 100 m. What is the distance between the man and the foot of the tower?**

**Solution:**

Given,

The height of the tower = 100 m and θ = 60°

Let the distance between the man and the foot of the tower = d

We have,

tan θ = Opposite side/Adjacent side

tan 60° = 100/d

√3 = 100/d [Since, tan 60° = √3]

d = 100/√3

Therefore, the distance between the man and the foot of the tower =

100/√3

**Example 6: Find the value of tan θ if sin θ = 7/25 and sec θ = 25/24.**

**Solution:**

Given,

sin θ = 7/25

sec θ = 25/24

We know that,

sec θ = 1/cos θ

25/24 = 1/cos θ cos θ = 24/25

We have,

tan θ = sin θ/cos θ

= (7/25)/(24/25)

= 7/24

Hence,

tan θ = 7/24

**Example 7: Find the value of tan θ if cosec θ = 5/3, and θ is the first quadrant angle.**

**Solution:**

Given, cosec θ = 5/3

From the Pythagorean identities, we have,

cosec

^{2}θ – cot^{2}θ = 1cot

^{2}θ = cosec^{2}θ – 1cot θ = (5/3)

^{2}– 1 = (25/9) – 1 = 16/9cot θ = ±√16/9 = ± 4/3

Since θ is the first quadrant angle, both cotangent and tangent functions are positive.

cot θ = 4/3

We know that,

cot θ = 1/tan θ

4/3 = 1/tan θ

tan θ = 3/4

Hence,

tan θ = 3/4

**Example 8: Find tan 3θ if sin θ = 3/7 and θ is the first quadrant angle.**

**Solution :**

Given, sin θ = 12/13

From the Pythagorean identities we have,

sin

^{2}θ + cos^{2}θ = 1cos

^{2}θ = 1 – sin^{2}θ = 1 – (12/13)^{2}cos2 θ = 1 – (144/169) = 25/169

cos θ = ±√25/169 = ±5/13

Since θ is the first quadrant angle, cos θ is positive.

cos θ = 5/13

We know that,

tan θ = sin θ/cos θ

= (12/25)/(5/13) = 12/5

Hence, tan θ = 12/5

Now, We know that ,

tan 3θ = (3 tan θ – tan3θ) / (1 – 3 tan2θ)tan 3θ = 3 × (12/5)

**Solved Example of Tangent Formula**

**Example: **Calculate the tangent angle of a right triangle whose adjacent side and opposite side are 8 cm and 6 cm respectively?

**Solution:**

Given,

Adjacent side (A)= 8 cm

Opposite side (O)= 6 cm

Using the formula of tangent:

**What is the Tangent Function?**

Tangent is the ratio of the opposite side divided by the adjacent side in a right-angled triangle. In trigonometry, there are six possible ratios. A ratio is a comparison of two numbers i.e. sides of a triangle. The Greek letter, θ, will be used to represent the reference angle in the right triangle. These six ratios are useful in different ways to compare two sides of a right triangle.

Tangent Angle Formula is normally useful to calculate the angle of the right triangle. In a right triangle, the tangent of an angle is a simple ratio of the length of the opposite side and the length of the adjacent side. Tangent is usually denoted as ‘tan’, but it is pronounced as a tangent. This function is useful to find out the length of a side of a triangle. It is possible when someone knows at least one side of the triangle and one of the acute angles.

The tangent function, along with sine and cosine functions, is one of the three most common trigonometric functions. Also in trigonometry, we may represent tan θ as the ratio of sin θ and cos θ.

**Solved Examples for ****Tangent Formula**

Q.1: Calculate the tangent angle of a right triangle whose adjacent side and opposite sides are 8 cm and 6 cm respectively?

Solution: Given,

Adjacent side i.e. base = 8 cm

Opposite side i.e. perpendicular = 6 cm

Also, the tangent formula is:

### Questions to be Solved

**Question 1**

Calculate the tangent angle of a right triangle whose adjacent side and opposite sides are 8 cm and 6 cm respectively?

Solution

Given, Adjacent side i.e. base = 8 cm

Opposite side i.e. perpendicular = 6 cm

Also, the tangent formula is: Tan θ=perpendicular/base

Tan θ=6/8 = 0.75

Therefore, tan θ=0.75

Thus the tangent value will be 0.75.

**Question 2**

Calculate the tangent angle of a right triangle whose adjacent side and opposite sides are 10 cm and 4 cm respectively?

Solution

Given, Adjacent side i.e. base = 10 cm

Opposite side i.e. perpendicular = 4 cm

Also, the tangent formula is: Tan θ=perpendicular/base

Tan θ=10/4 = 2.5

Therefore, tan θ= 2.5

Thus the tangent value will be 2.5.

## FAQs on Tangent Formula

**1. What is Tan Equal to?**

The tangent of x is defined to be its sine divided by its cosine: tan x = sin x/ cos x. The cotangent of x can be defined to be the cosine of x divided by the sine of x: cot x = cos x/ sin x.

**2. What is the Symbol for a Tangent?**

The ratio of the adjacent side of a right triangle to the hypotenuse is called the cosine and given the symbol cos. Now finally, the ratio of the opposite side to the adjacent side is known as the tangent and given the symbol tan.

**3. Where is Tangent 1?**

The exact value of arctan(1) is equal to

. The tangent function is known to be positive in the first and third quadrants. Now to find the second solution, we need to add the reference angle from π to find the solution in the fourth quadrant.

**4. What is the Symbol for Pi?**

Pi is denoted by the symbol π and is defined as the ratio of the circumference of a circle to its diameter, pi, or in symbol form, π, which seems a simple enough concept. But it rather turns out to be an “irrational number,” which means that its exact value is inherently unknowable.

**5. What are the applications of tangent formulas?**

The tangent formulae talk about the tangent function also called the tan function. In order to understand this formula, you can consider a right-angled triangle with one of its acute angles as x. The tangent formula will then be tan x = (opposite side) / (adjacent side), where the opposite side will be the side opposite to angle x and the adjacent side will be the side adjacent to angle x. This formula has various forms. These forms can be used to check proof of a trigonometric identity or to find out the value of the tangent function with the information that is available.

**6. What is the origin of trigonometric ratios?**

Greek words trigonal which means “triangle” and metro on which means “to measure” when mixed together produce the word Trigonometry. It was until the 16th-century that trigonometry was highly concerned with the computation of the numerical values of the missing parts of a triangle for which the values for the other parts present are provided. For example, if there are lengths of two sides that are provided by a triangle then the two remaining angles can be calculated. These calculations tend to differentiate trigonometry from geometry.

**7. How many tangent formulae are present?**

### There are a total of 8 tangent formulae that are present and can be provided as follows:

1. Tangent formulas using reciprocal identity

2. Tangent formula using sin and cos

3. Tangent formulas using Pythagorean identity

4. Tangent formulas using cofunction identities

5. Tangent formulas using sum/difference formulas

6. Tangent formulas of double angle

7. The tangent formula of triple angle

8. The tangent formula of half-angle

All of these tangent formulae can be used for various determination of functions.

## Formula

### General Formula:

(Based on the sides of trigonometry)

tan = Perpendicular/Base

(Respective to the angle x on the right-angled triangle)

tan = opposite side / adjacent side

### Sin and cos Tangent Formula:

tan x = (sin x) / (cos x)

### Tangent Formulas Using Reciprocal Identity:

tan x = 1 / (cot x)

### Tangent Formulas Using Pythagorean Identity:

tan x = ± √( sec^{2}x – 1)

### Cofunction Identity Tangent Formula:

tan x = cot (90° – x) (OR)

tan x = cot (π/2 – x)

### Sum and difference Tangent Formula:

tan (A – B) = (tan A – tan B) / (1 + tan A tan B)

tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

### Double angle Tangent Formula:

tan 2x = (2 tan x) / (1 – tan^{2}x)

### Triple Angle Tangent Formula:

tan 3x = (3 tan x – tan^{3}x) / (1 – 3tan^{2}x)

### Half Angle Tangent Formula:

tan (x/2) = (1 – cos x) / ( sin x)

tan (x/2) =± √[(1 – cos x) / (1 + cos x)]

## Solved Examples

**1. How is the cos sin tangent formula derived? **

According to the ratios in trigonometry:

sin x = (perpendicular) / (hypotenuse)

cos = (base) / (hypotenuse)

tan = Perpendicular/base

(sin x) / (cos x) = [ (perpendicular) / (hypotenuse) ] / [ (base) / (hypotenuse) ]

tan x = Perpendicular/Base

**2. How to derive a tangent formula using Pythagoras theorem? **

The tangent formula is derived from the Pythagoras theorem, by using the identity:

sec^{2}x – tan^{2}x = 1

On subtracting 1 from both sides in the given identity:

-tan^{2}x = 1 – sec^{2}x

On performing the product of the whole expression with – 1.

tan^{2}x = sec^{2}x – 1

Taking the square root on both sides:

tan x = ± √( sec^{2}x – 1)

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