Examples 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,

sin² θ + cos² θ = 1

cos² θ = 1 – sin² θ = 1 – (2/5)²

cos² 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

= (2/5)/(√21/5) = 2/√21

tan θ = 2√21 /21

So, value of tan θ when sin θ = 2/5 and θ is in first quadrant is (2√21) /(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,

sec² x – tan² x = 1

tan² x = sec² x – 1= (13/12)² – 1

tan² 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,

Height of tower = 100 m and θ = 60°

Let distance between man and foot of 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, distance between the man and the foot of tower is 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,

cosec² θ – cot² θ = 1

cot² θ = cosec² 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,

sin²θ + cos²θ = 1

cos²θ = 1 – sin²θ = 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)

Tangent Function is among the six basic trigonometric functions and is calculated by taking the ratio of the perpendicular side and the hypotenuse side of the right-angle triangle.

In this article, we will learn about Trigonometric ratios, Tangent formulas, related examples, and others in detail.