Tangent Formulas
Trigonometry is an important branch of mathematics that deals with the relationship between angles and lengths of the sides of a right-angled triangle. The six trigonometric ratios or functions are sine, cosine, tangent, cosecant, and secant, and a trigonometric ratio is a ratio between the sides of a right-angled triangle. Sine, cosine, and tangent functions are three important trigonometric functions since the other three, i.e., cosecant, secant, and cotangent functions are the reciprocal functions of sine, cosine, and tangent functions, respectively.
- sin θ = Opposite side/Hypotenuse
- cos θ = Adjacent side/Hypotenuse
- tan θ = Opposite side/Adjacent side
- cosec θ = Hypotenuse/Opposite side
- sec θ = Hypotenuse/Adjacent side
- cot θ = Adjacent side/Opposite side
Tangent Formula
The tangent of an angle in a right-angled triangle is the ratio of the length of the opposite side to the length of the adjacent side to the given angle. We write a tangent function as “tan”. Let us consider a right-angled triangle XYZ and one of its acute angles is “θ”. An opposite side is the side that is opposite to the angle “θ” and the adjacent side is the side that is adjacent to the angle “θ”.
Now, the tangent formula for the given angle “θ” is,
tan θ = Opposite side/Adjacent side
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 θ
We 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 = 1
cos^{2} θ = 1 – sin^{2} θ
cos 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 θ
We know that,
tan θ = sin θ/cos θ
From the Pythagorean identities, we have,
sin^{2} θ + cos^{2} θ = 1
sin^{2} θ = 1 – cos^{2} θ
sin 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 θ
or
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 = 1
cot^{2} θ = cosec^{2 }θ – 1
cot 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 – 1
From the Pythagorean identities, we have,
sec^{2} θ – tan^{2} θ = 1
tan θ = sec^{2} θ – 1
Hence, 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 (In degrees) |
Angle (In Radians) |
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 = 1
cos^{2} θ = 1 – sin^{2} θ = 1 – (2/5)^{2}
cos^{2} θ = 1 – (4/5) = 21/25
cos 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 = 1
tan^{2} x = sec^{2} x – 1= (13/12)^{2 }– 1
tan^{2} x = (169/144) – 1= 25/144
tan 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 = 1
cot^{2} θ = cosec^{2} θ – 1
cot θ = (5/3)^{2} – 1 = (25/9) – 1 = 16/9
cot 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 = 1
cos^{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)
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