Compute the exact value of tan (150° + 45°)

Trigonometry is the branch of mathematics that relates the sides and angles in a triangle using their ratios. Trigonometry helps in calculating various measurements connected to a triangle. In Trigonometry, standard ratios are defined for the ease of calculation of some common problems related to the length and angles of the sides of a right-angled triangle.

Trigonometric Ratios

A trigonometric ratio is the proportion of sides with either of the acute angles in the right-angled triangle. It can be defined as a simple trigonometric ratio in terms of sides of a right-angled triangle i.e. the hypotenuse, base side, and the perpendicular side. There are three standard trigonometric ratios wiz. sine, cosine, and tangent.

• Sine is the function that takes in the parameter an angle θ, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the opposite side to the hypotenuse of the right-angled triangle. In technical terms, it can be written as,

sin(θ) = opposite side/hypotenuse

• Cosine is the function that takes in the parameter an angle θ, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the adjacent side to the hypotenuse of the right-angled triangle. In technical terms, it can be written as,

cos(θ) = adjacent side/hypotenuse

• Tangent is the function that takes in the parameter an angle θ, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the opposite side to the adjacent side of the right-angled triangle. In technical terms, it can be written as,

tan(θ) = opposite side / adjacent side

Trigonometric Table

The below is the table for some common angles and the basic trigonometric ratios. The value of each angle in the trigonometry is fixed and known but the ones mentioned are more common and mostly used,

Ratio\Angle	0°	30°	45°	60°	90°
sin(θ)	0	1/2	1/√2	√3/2	1
cos(θ)	1	√3/2	1/√2	1/2	0
tan(θ)	0	1/√3	1	√3	∞
cosec(θ)	∞	2	√2	2/√3	1
sec(θ)	1	2/√3	√2	2	∞
cot(θ)	∞	√3	1	1/√3	0

There are also some other trigonometric ratios to apply beyond the right-angled triangles:

1. sin(-θ) = – sin(θ)
2. cos(-θ) = cos(θ)
3. tan(-θ) = – tan(θ)

For this problem, look at certain tangent ratio-specific formulae and relations to things simple and easy to understand. Look into the Complementary and Supplementary angles of the tangent function,

Complementary and Supplementary angles

Complementary angles are a pair of angles that add up to form 90° or π/2 radians. Such angles can be formed and find the equivalent angles in terms of the trigonometric ratios. 

Supplementary angles are a pair of angles that add up to form 180° or π radians. Such angles can be formed and find the equivalent angles in terms of the trigonometric ratios. 

Subtract an angle from 90° to obtain a pair of complementary angle and similarly, one can add up an angle to 90°  to form a supplementary angle pair. In other words, the actual angle can be adjusted in the function of trigonometric ratios to form either complementary or supplementary angles and then evaluate the deduced trigonometric ratio as per the list of formulas given below.

• tan(nπ/2 + θ) = -cot(θ) or tan(n × 90° + θ) = -cot(θ)
• tan(nπ/2 – θ) = cot(θ) or tan(n × 90° – θ) = cot(θ)
• tan(nπ + θ) = tan(θ) or tan(n × 180° + θ) = tan(θ)
• tan(nπ – θ) = -tan(θ) or sin(n × 180° – θ) = -tan(θ)
• tan(3nπ/2 – θ) = cot(θ) or tan(n × 270° + θ) = cot(θ)
• tan(3nπ/2 + θ) = -cot(θ) or tan(n × 270° + θ) = -cot(θ)
• tan(2nπ + θ) = tan(θ) or tan(n × 360°+ θ) = tan(θ)
• tan(2nπ – θ) = -tan(θ) or tan(n × 360° – θ) = -tan(θ)

There are Compound angles formula for trigonometry.

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

Find the value of tan(150° + 45°)

Solution:

Method 1

tan(150° + 45°) 

Use the compound angles formula for tangent function,

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

So, here, A = 150° and B = 45°,

tan(150° + 45°) = (tan(150°) + tan(45°)) / [1 – (tan(150).tan(45))]

Now, write 150° as (180° – 30°),

So, tan(150°) = tan(180° – 30°)

sin(n × 180° – θ) = -tan(θ)

Here n = 1 and θ = 30

So,

tan(150°) = tan(180° – 30°)

= -tan(30°)

= – 1/√3

Thus,

tan(150° + 45°) = ( tan(150°) + tan(45°) ) / [1 – (tan(150).tan(45))]

tan(150°) = -1/√3 and tan(45°) = 1,

= ((-1/√3) + 1) / [1 – ((-1/√3).(1))]

= [(-1 + √3)/√3] / [1 + 1/√3]

= [(√3 – 1)/√3] / [(√3 + 1)/√3]    

= [√3 – 1] / [√3 +1]

Rationalizing the denominator (multiplying and dividing by the denominator’s conjugate)

tan(150° + 45°) = [(√3 – √1) × (√3 – 1)] / [(√3 + 1) × (√3 – 1)]

= [(√3 – 1)²] / [ (√3²) – (1²)]

= [3 – 2√3 + 1 ] / [3 – 1]

= [4 – 2√3] / 2

= 2 – √3

~ 0.268

Thus the value of tan(150° + 45°) = 2 – √3 ~ 0.268

Method 2

tan(150° + 45°)

= tan(195°)

Now, write 195° as (180° + 15°),

So, tan(195°) = tan(180° + 45°)

sin(n × 180° + θ) = tan(θ)

Here, n = 1 and θ = 15

tan(150° + 45°) = tan(195°) = tan(180° +15°)

= tan(15°)

Now, 15° can be written as (45° – 30°)

Thus,

tan(150° + 45°) = tan(15°)

= tan(45° – 30°)

The compound angles formula for tangent function,

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

Here, A = 45° and B = 30°

tan(45° – 30°) = (tan(45°) – tan(30°)) / [1 + (tan(45).tan(30))]

tan(45°) = -1 and tan(30°) = 1/√3,

= (1 – (1/√3)) / [1 + ((1).(1/√3))]

= [(1 – √3)/√3] / [1 + 1/√3]

= [(√3 – 1)/√3] / [(√3 + 1)/√3]    

= [√3 – 1] / [√3 +1]

tan(150° + 45°) = [√3 – 1] / [√3 +1]

Rationalizing the denominator (multiplying and dividing by the denominator’s conjugate)

tan(150° + 45°) = tan(15°)

= [(√3 – √1) × (√3 – 1)] / [(√3 + 1) × (√3 – 1)]

= [(√3 – 1)²] / [ (√3²) – (1²)]

= [3 – 2√3 + 1] / [3 – 1]

= [4 – 2√3] / 2

= 2 – √3

~ 0.268

Thus the value of tan(150° + 45°) = 2 – √3 ~ 0.268

Method 3

tan(150° + 45°)

= tan(195°)

Now, write 195° as (240° – 45°),

So, tan(195°) = tan(240° – 45°)

Use the compound angles formula for tangent function,

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

Here. A = 240° and B = 45°,

tan(240° – 45°) = (tan(240°) – tan(45°)) / [1 + (tan(240).tan(45))]

Now, write 240° as (270° – 30°),

So, tan(240°) = tan(270° – 30°)

sin(n × 270° – θ) = cot(θ)

Here n = 1 and θ = 30,

So, tan(240°) = tan(270° – 30°)

= cot(30°)

= √3

Hence , tan(240°) = √3

tan(150° + 45°) = tan(195°) = tan(240° – 45°)

tan(240° – 45°) = (tan(240°) – tan(45°)) / [1 – (tan(240).tan(45))]

Here, tan(240°) = 3 and tan(45°) = 1

= (tan(240°) – tan(45°)) / [1 + (tan(240°).tan(45°))]

= (3 – 1) / (1 + (3).(1))

tan(150° + 45°) = [√3-1] / [√3 +1]

Rationalizing the denominator (multiplying and dividing by the denominator’s conjugate)

= [(√3 – 1)²] / [ (√3²) – (1²)]

= [3 – 2√3 + 1] / [3 – 1]

= [4 – 2√3] / 2

= 2 – √3

~ 0.268

Thus the value of tan(150° + 45°) = 2 – √3 ~ 0.268

Sample Problems

Question 1: Find the value of tan(135°)

Solution: 

tan(135°) 

135° as (90° + 45°)

So,

tan(135°) = tan(90° + 45°)

 tan(n × 90° + θ) = -cot(θ)

Here, n = 1 and θ = 45°

Thus,

tan(135°) = tan(90° + 45°)

= -cot(45)

= -1

Thus, the value of tan(135°) = -1

Question 2: Find the value of tan(210°)

Solution: 

tan(210°)

210° as (180° + 30°)

So,

tan(210°) = tan(180° + 30°)

tan(n × 180° + θ) = tan(θ)

Here, n = 1 and  θ = 30°,

Thus,

tan(210°) = tan(180° + 30°)

= tan(30°)

= 1/√3

Thus,

tan(210°) = 1/√3

Thus, the value of tan(210°) = 1/√3 ~ 0.5773

Question 3: Find the value of tan(135° + 30°)

Solution:

tan(135° + 30°)

Use the compound angles formula for tangent function,

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

Here. A = 135° and B = 30°,

tan(135° + 30°) = [tan(135°) + tan(30°)] / [1 – ((tan(135°).(tan(30°))]

To find, tan(135°),

Write 135° as (90° + 45°)

So, tan(135°) = tan(90° + 45°)

tan(n × 90° + θ) = -cot(θ)

Here, n = 1 and θ = 45°

So, tan(135°) = tan(90° + 45°)

= -cot(45°)

= -1

tan(135°) = -1

Thus,

tan(135° + 30°) = [tan(135°) + tan(30°)] / [1 – ((tan(135°).(tan(30°))]

Here, tan(135°) = -1 and tan(30°) = 1/√3,

tan(135° + 30°) = [tan(135°) + tan(30°)] / [1 – ((tan(135°).(tan(30°))]

= [-1 + (1/√3)] / [1 – ((-1).(1/√3))]

= [(-√3 +1) / √3] / [1 + (1/√3)]

= [(-√3 + 1) / √3] / [(√3 + 1) / √3] 

= [-√3 + 1] / [√3 +1]

tan(135° + 30°) = [-√3 + 1] / [√3 + 1]

Rationalizing the denominator (multiplying and dividing by the denominator’s conjugate)

= [(-√3 + 1) × (√3 – 1)] / [(√3 + 1) × (√3 – 1)]

= [-√3 × √3 + √3 + √3 – 1] / [ (√3²) – (1²) ]

= [-3 + 2√3 – 1] / [3 – 1]

= [-4 + 2√3 ] /2 

= -2 + √3

~ -0.268

 Thus, the value of tan(135° + 30°) = -2 + √3 ~ -0.268