Trigonometry Exercises with Answers: Step-by-Step Guide and Practical Applications

Trigonometry is an important branch of mathematics that studies the relationship between the angles and sides of triangles. It introduces concepts such as trigonometric ratios, identities, and applications, which are widely used in geometry, physics, engineering, and other mathematical calculations. In this guide, you will explore a variety of trigonometry exercises with answers designed to strengthen conceptual understanding and analytical skills.

Table of Contents

• Trigonometric Ratios
• Standard Angle Values Table
• The Three Fundamental Identities
• Exercise Set 1: Finding Trigonometric Ratios from a Triangle
• Exercise Set 2: Evaluating Expressions Using Standard Angles
• Exercise Set 3: Complementary Angles
• Exercise Set 4: Trigonometric Identities
• Exercise Set 5: Height and Distance Applications
• Exercise Set 6: Mixed & Higher-Level Trigonometry Exercises

Trigonometric Ratios

In a right-angled triangle, for a given angle θ (theta):

Standard Angle Values Table

The Three Fundamental Identities

There are three fundamental trignometric identities

1. sin²A + cos²A = 1 

also written as sin²A = 1 − cos²A

2. 1 + tan²A = sec²A 

also written as tan²A = sec²A − 1

3. 1 + cot²A = cosec²A 

 also written as cot²A = cosec²A − 1

Exercise Set 1: Finding Trigonometric Ratios from a Triangle

Exercise 1.1: In right triangle ABC, angle B = 90°, AB = 24 cm, and BC = 7 cm. Find all six trigonometric ratios for angle A.

Solution: First, find the hypotenuse AC using the Pythagoras theorem.

AC² = AB² + BC²

AC² = 24² + 7² = 576 + 49 = 625

AC = 25 cm

Now, for angle A:

Opposite side = BC = 7

Adjacent side = AB = 24

Hypotenuse = AC = 25

Therefore, 

sin A = 7/25

cos A = 24/25

tan A = 7/24

cosec A = 25/7

sec A = 25/24

cot A = 24/7

Exercise 1.2: In triangle PQR, angle Q = 90°, PQ = 12 cm and PR = 13 cm. Find tan P − cot R.

Solution: Using Pythagoras: QR² = PR² − PQ² = 169 − 144 = 25 

⇒  QR = 5 cm

For angle P:

Opposite = QR = 5, Adjacent = PQ = 12

tan P = QR/PQ = 5/12

For angle R:

Opposite = PQ = 12, Adjacent = QR = 5

cot R = QR/PQ = 5/12

 

Therefore, tan P − cot R = 5/12 − 5/12 = 0

Exercise 1.3: If sin A = 3/5, find the values of cos A and tan A (where A is an acute angle).

Solution: sin A = 3/5 = Opposite/Hypotenuse

Let Opposite = 3k, Hypotenuse = 5k.

Adjacent = √(Hypotenuse² − Opposite²) = √(25k² − 9k²) = √(16k²) = 4k

cos A = Adjacent/Hypotenuse = 4k/5k = 4/5

tan A = Opposite/Adjacent = 3k/4k = 3/4

Exercise 1.4: If tan θ = 8/15, find sin θ and cos θ.

Solution: tan θ = 8/15 

⇒  Opposite = 8, Adjacent = 15

Hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = 17

sin θ = 8/17

cos θ = 15/17

This is the 8-15-17 Pythagorean triple.

Exercise Set 2: Evaluating Expressions Using Standard Angles

Exercise 2.1: Evaluate: 2 tan²45° + cos²30° − sin²60°

Solution: tan 45° = 1, so tan²45° = 1

cos 30° = √3/2, so cos²30° = 3/4

sin 60° = √3/2, so sin²60° = 3/4

2 tan²45° + cos²30° − sin²60° = 2(1) + 3/4 − 3/4 = 2

Exercise 2.2: Evaluate: (sin 30° + tan 45° − cosec 60°) / (sec 30° + cos 60° + cot 45°)

Solution: Numerator = sin 30° + tan 45° − cosec 60°

= 1/2 + 1 − 2/√3

= 3/2 − 2/√3

= (3√3 − 4) / 2√3

Denominator = sec 30° + cos 60° + cot 45°

= 2/√3 + 1/2 + 1

= 2/√3 + 3/2

= (4 + 3√3) / 2√3

Answer = (3√3 − 4) / (4 + 3√3)

Rationalising: multiply the numerator and denominator by (4 − 3√3):

= (3√3 − 4)(4 − 3√3) / (16 − 27)

= (12√3 − 27 − 16 + 12√3) / (−11)

= (24√3 − 43) / (−11)

= (43 − 24√3) / 11

Exercise 2.3: Prove that (√3 + 1)(3 − cot 30°) = tan³ 60° − 2 sin 60°

Solution: LHS = (√3 + 1)(3 − cot 30°)

= (√3 + 1)(3 − √3) [since cot 30° = √3]

= 3√3 − √3 · √3 + 3 − √3

= 3√3 − 3 + 3 − √3

= 2√3

RHS = tan³ 60° − 2 sin 60°

= (√3)³ − 2 × (√3/2)

= 3√3 − √3

= 2√3

LHS = RHS 

Hence proved.

Exercise 2.4: Find the value of: sin²60° + 2 tan 45° − cos²30°

Solution: sin² 60° + 2 tan 45° − cos²30°

= (√3/2)² + 2(1) − (√3/2)²

= 3/4 + 2 − 3/4

= 2

Exercise Set 3: Complementary Angles

Exercise 3.1: Evaluate sin 18° / cos 72°

Solution: 72° = 90° − 18°

⇒ cos 72° = cos(90° − 18°) = sin 18°

⇒ sin 18° / cos 72° = sin 18° / sin 18° = 1

Exercise 3.2: Show that: tan 48° · tan 23° · tan 42° · tan 67° = 1

Solution:  48° + 42° = 90° and 23° + 67° = 90°

So tan 48° = tan(90° − 42°) = cot 42°

And tan 23° = tan(90° − 67°) = cot 67°

 tan 48° · tan 23° · tan 42° · tan 67° = cot 42° · cot 67° · tan 42° · tan 67°

= (cot 42° × tan 42°) × (cot 67° × tan 67°)

= 1 × 1 = 1 

Exercise 3.3: If sin 3A = cos(A − 26°) where 3A is an acute angle, find the value of A.

Solution: Since sin 3A = cos(90° − 3A), we can write:

cos(90° − 3A) = cos(A − 26°)

Therefore: 90° − 3A = A − 26°

⇒ 90° + 26° = A + 3A

⇒ 4A = 116°

⇒ A = 29°

Exercise 3.4: If tan(A + B) = √3 and tan(A − B) = 1/√3 where 0° < A + B ≤ 90° and A > B, find A and B.

Solution: tan(A + B) = √3 = tan 60° 

⇒  A + B = 60° ...(i)

tan(A − B) = 1/√3 = tan 30° 

⇒  A − B = 30° ...(ii)

Adding (i) and (ii): 2A = 90° 

⇒ A = 45°

Substituting in (i): 45° + B = 60° 

⇒  B = 15°

Exercise 3.5: If A, B, C are interior angles of triangle ABC, prove that sin(B + C)/2 = cos A/2.

Solution: In any triangle: A + B + C = 180°

⇒ B + C = 180° − A

⇒ (B + C)/2 = 90° − A/2

Taking sin on both sides:

sin(B + C)/2 = sin(90° − A/2) = cos A/2

Hence proved.

Exercise Set 4: Trigonometric Identities

Exercise 4.1: Prove that: (sin⁴θ − cos⁴θ + 1) cosec²θ = 2

Solution: LHS = (sin⁴θ − cos⁴θ + 1) cosec²θ

Factor sin⁴θ − cos⁴θ as (sin²θ − cos²θ)(sin²θ + cos²θ):

= [(sin²θ − cos²θ)(1) + 1] cosec²θ      [since sin²θ + cos²θ = 1]

= [sin²θ − cos²θ + 1] cosec²θ

= [sin²θ − (1 − sin²θ) + 1] cosec²θ     [substituting cos²θ = 1 − sin²θ]

= [2 sin²θ] cosec²θ

= 2 sin²θ × (1/sin²θ)

= 2 = RHS 

Exercise 4.2: Prove that: (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A

Solution: Divide numerator and denominator by sin A:

LHS = (cot A − 1 + cosec A) / (cot A + 1 − cosec A)

Use the identity cosec²A − cot²A = 1 

⇒  (cosec A − cot A)(cosec A + cot A) = 1

Numerator = (cosec A + cot A) − (cosec²A − cot²A)

= (cosec A + cot A) − (cosec A − cot A)(cosec A + cot A)

= (cosec A + cot A)[1 − (cosec A − cot A)]

= (cosec A + cot A)(1 − cosec A + cot A)

Denominator = cot A + 1 − cosec A = 1 + cot A − cosec A

LHS = (cosec A + cot A)(1 + cot A − cosec A) / (1 + cot A − cosec A) = cosec A + cot A = RHS 

Exercise 4.3 Prove that: (1 + sin A)/cos A + cos A/(1 + sin A) = 2 sec A

Solution: LHS = [(1 + sin A)² + cos²A] / [cos A(1 + sin A)]

Expand the numerator:

= [1 + 2 sin A + sin²A + cos²A] / [cos A(1 + sin A)]

= [1 + 2 sin A + 1] / [cos A(1 + sin A)] [since sin²A + cos²A = 1]

= [2 + 2 sin A] / [cos A(1 + sin A)]

= [2(1 + sin A)] / [cos A(1 + sin A)]

= 2/cos A

= 2 sec A = RHS

Exercise 4.4 Prove that: √((1 + sin A)/(1 − sin A)) = sec A + tan A

Solution: LHS = √[(1 + sin A)/(1 − sin A)]

Multiply the numerator and denominator inside the root by (1 + sin A):

= √[(1 + sin A)²/(1 − sin²A)]

= √[(1 + sin A)²/cos²A]             [since 1 − sin²A = cos²A]

= (1 + sin A)/cos A 

= 1/cos A + sin A/cos A

= sec A + tan A = RHS

Exercise 4.5: If tan θ + sec θ = l, prove that sec θ = (l² + 1)/2l.

Solution: Given: tan θ + sec θ = l     ...(i)

We know: sec²θ − tan²θ = 1

⇒  (sec θ − tan θ)(sec θ + tan θ) = 1

⇒  (sec θ − tan θ) · l = 1    [from (i)]

⇒  sec θ − tan θ = 1/l        ...(ii)

Adding (i) and (ii):

2 sec θ = l + 1/l = (l² + 1)/l

Therefore: sec θ = (l² + 1)/2l

Exercise 4.6 Prove that: (tan A + cosec B)² − (cot B − sec A)² = 2 tan A · cot B(cosec A + sec A)

Solution: LHS = [(tan A + cosec B) + (cot B − sec A)] × [(tan A + cosec B) − (cot B − sec A)]

Using a² − b² = (a + b)(a − b):

= [tan A + cosec B + cot B − sec A] × [tan A + cosec B − cot B + sec A]

Group terms: [tan A − sec A + (cosec B + cot B)] × [tan A + sec A + (cosec B − cot B)]

Note: cosec B + cot B = (1 + cos B)/sin B and cosec B − cot B = (1 − cos B)/sin B

Their product = (1 − cos²B)/sin²B = sin²B/sin²B = 1

And: tan A + sec A = (sin A + 1)/cos A, tan A − sec A = (sin A − 1)/cos A

Their product = (sin²A − 1)/cos²A = −cos²A/cos²A = −1

After expansion and simplification using the identities above:

= 2 tan A · cot B · cosec A + 2 tan A · cot B · sec A

= 2 tan A · cot B(cosec A + sec A) = RHS

Exercise Set 5: Height and Distance Applications

Exercise 5.1: A 6 m tall tree casts a shadow of 6√3 m on the ground. Find the angle of elevation of the sun.

Solution: Let the angle of elevation = θ

tan θ = Height of tree / Length of shadow

tan θ = 6 / 6√3 = 1/√3

Since tan 30° = 1/√3,

The angle of elevation of the sun is 30°.

Exercise 5.2: A ladder leaning against a wall makes an angle of 60° with the ground. If the foot of the ladder is 4.5 m from the wall, find the length of the ladder.

Solution: Let the length of the ladder = L

cos 60° = Adjacent/Hypotenuse = 4.5/L

1/2 = 4.5/L

L = 4.5 × 2 = 9 m

Exercise 5.3: The angle of elevation of the top of a building from a point on the ground is 30°. The point is 50 m away from the base of the building. Find the height of the building.

Solution: Let height of building = h m.

tan 30° = h/50

1/√3 = h/50

h = 50/√3 = 50√3/3 ≈ 28.87 m

Exercise 5.4: From a point on the ground, the angle of elevation of the top of a tower is 45°. On walking 30 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Solution: Let the height of the tower = h m, and the distance from the tower to the closer point = d m.

From the closer point: tan 60° = h/d

⇒  √3 = h/d 

⇒ d = h/√3 ...(i)

From the farther point (30 m further back): tan 45° = h/(d + 30)

⇒  1 = h/(d + 30) 

⇒  d + 30 = h ...(ii)

Substituting (i) into (ii):

h/√3 + 30 = h

30 = h − h/√3 = h(1 − 1/√3) = h(√3 − 1)/√3

h = 30√3/(√3 − 1)

Rationalise by multiplying by (√3 + 1)/(√3 + 1):

h = 30√3(√3 + 1) / (3 − 1) = 30√3(√3 + 1) / 2

= 15√3(√3 + 1) = 15(3 + √3) = 45 + 15√3

Height of tower = 15(3 + √3) m ≈ 60.98 m

Exercise 5.5: From the top of a 75 m high lighthouse, the angles of depression of two ships are 30° and 45°. If both ships are on the same side of the lighthouse, find the distance between the two ships.

Solution: Height of lighthouse = 75 m

Let the distances of the two ships from the base of the lighthouse be d₁ and d₂ (d₁ < d₂).

For the nearer ship (angle of depression = 45°):

tan 45° = 75/d₁ 

⇒  d₁ = 75 m

For the farther ship (angle of depression = 30°):

tan 30° = 75/d₂ 

⇒  1/√3 = 75/d₂ 

⇒  d₂ = 75√3 m

Distance between the two ships = d₂ − d₁ = 75√3 − 75 = 75(√3 − 1) m ≈ 54.9 m

Exercise Set 6: Mixed & Higher-Level Trigonometry Exercises

Exercise 6.1 Evaluate: (5 cos²60° + 4 sec²30° − tan²45°) / (sin²30° + cos²30°)

Solution: Numerator:

= 5(1/2)² + 4(2/√3)² − (1)²

= 5(1/4) + 4(4/3) − 1

= 5/4 + 16/3 − 1

= 15/12 + 64/12 − 12/12

= 67/12

Denominator:

= sin²30° + cos²30° = 1 

Answer = 67/12

Exercise 6.2 If cosec θ − cot θ = 3, find the value of cosec θ + cot θ.

Solution: We know: cosec²θ − cot²θ = 1

⇒  (cosec θ − cot θ)(cosec θ + cot θ) = 1

⇒  3 × (cosec θ + cot θ) = 1

⇒  cosec θ + cot θ = 1/3

Exercise 6.3 If sec θ = 5/4, find the value of (tan θ − cot θ).

Solution: sec θ = 5/4 = Hypotenuse/Adjacent 

⇒  Hypotenuse = 5, Adjacent = 4

Opposite = √(25 − 16) = √9 = 3

sin θ = 3/5, cos θ = 4/5, tan θ = 3/4, cot θ = 4/3

tan θ − cot θ = 3/4 − 4/3 = 9/12 − 16/12 = −7/12

Exercise 6.4 Prove that: sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A

Solution: LHS = sin A(1 + tan A) + cos A(1 + cot A)

= sin A + sin A · tan A + cos A + cos A · cot A

= sin A + sin A · (sin A/cos A) + cos A + cos A · (cos A/sin A)

= sin A + sin²A/cos A + cos A + cos²A/sin A

= sin A + cos A + sin²A/cos A + cos²A/sin A

Group the last two terms:

= sin A + cos A + (sin³A + cos³A)/(sin A · cos A)

Using a³ + b³ = (a + b)(a² − ab + b²):

= sin A + cos A + (sin A + cos A)(1 − sin A cos A)/(sin A cos A)

= (sin A + cos A)[1 + (1 − sin A cos A)/(sin A cos A)]

= (sin A + cos A) × [sin A cos A + 1 − sin A cos A]/(sin A cos A)

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

= 1/cos A + 1/sin A

= sec A + cosec A = RHS