Trigonometry Questions

Trigonometry questions given here involve finding the missing sides of a triangle with the help of trigonometric ratios and proving trigonometry identities. We know that trigonometry is one of the most important chapters of Class 10 Maths. Hence, solving these questions will help you to improve your problem-solving skills.

What is Trigonometry?

The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the sides and angles of a triangle.

The basic trigonometric ratios are defined as follows.

sine of ∠A = sin A = Side opposite to ∠A/ Hypotenuse

cosine of ∠A = cos A = Side adjacent to ∠A/ Hypotenuse

tangent of ∠A = tan A = (Side opposite to ∠A)/ (Side adjacent to ∠A)

cosecant of ∠A = cosec A = 1/sin A = Hypotenuse/ Side opposite to ∠A

secant of ∠A = sec A = 1/cos A = Hypotenuse/ Side adjacent to ∠A

cotangent of ∠A = cot A = 1/tan A = (Side adjacent to ∠A)/ (Side opposite to ∠A)

Also, tan A = sin A/cos A

cot A = cos A/sin A

Also, read: Trigonometry

Trigonometry Questions and Answers

1. From the given figure, find tan P – cot R.

Solution:

From the given,

PQ = 12 cm

PR = 13 cm

In the right triangle PQR, Q is right angle.

By Pythagoras theorem,

PR² = PQ² + QR²

QR² = (13)² – (12)²

= 169 – 144

= 25

QR = 5 cm

tan P = QR/PQ = 5/12

cot R = QR/PQ = 5/12

So, tan P – cot R = (5/12) – (5/12) = 0

2. Prove that (sin⁴θ – cos⁴θ +1) cosec²θ = 2

Solution:

L.H.S. = (sin⁴θ – cos⁴θ +1) cosec²θ

= [(sin²θ – cos²θ) (sin²θ + cos²θ) + 1] cosec²θ

Using the identity sin²A + cos²A = 1,

= (sin²θ – cos²θ + 1) cosec²θ

= [sin²θ – (1 – sin²θ) + 1] cosec²θ

= 2 sin²θ cosec²θ

= 2 sin²θ (1/sin²θ)

= 2

= RHS

3. Prove that (√3 + 1) (3 – cot 30°) = tan³60° – 2 sin 60°.

Solution:

LHS = (√3 + 1)(3 – cot 30°)

= (√3 + 1)(3 – √3)

= 3√3 – √3.√3 + 3 – √3

= 2√3 – 3 + 3

= 2√3

RHS = tan³60° – 2 sin 60°

= (√3)³ – 2(√3/2)

= 3√3 – √3

= 2√3

Therefore, (√3 + 1) (3 – cot 30°) = tan³60° – 2 sin 60°.

Hence proved.

4. If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

Solution:

tan(A + B) = √3

tan(A + B) = tan 60°

A + B = 60°….(i)

And

tan(A – B) = 1/√3

tan(A – B) = tan 30°

A – B = 30°….(ii)

Adding (i) and (ii),

A + B + A – B = 60° + 30°

2A = 90°

A = 45°

Substituting A = 45° in (i),

45° + B = 60°

B = 60° – 45° = 15°

Therefore, A = 45° and B = 15°.

5. If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

Solution:

Given,

sin 3A = cos(A – 26°); 3A is an acute angle

cos(90° – 3A) = cos(A – 26°) {since cos(90° – A) = sin A}

⇒ 90° – 3A = A – 26

⇒ 3A + A = 90° + 26°

⇒ 4A = 116°

⇒ A = 116°/4

⇒ A = 29°

6. If A, B and C are interior angles of a triangle ABC, show that sin (B + C/2) = cos A/2.

Solution:

We know that, for a given triangle, the sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

B + C = 180° – A

Dividing both sides of this equation by 2, we get;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Take sin on both sides,

sin (B + C)/2 = sin (90° – A/2)

⇒ sin (B + C)/2 = cos A/2 {since sin(90° – x) = cos x}

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

Solution:

Given,

tan θ + sec θ = l….(i)

We know that,

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),

tan θ + sec θ + sec θ – tan θ = l + (1/l)

2 sec θ = (l² + 1)l

sec θ = (l² + 1)/2l

Hence proved.

8. Prove that (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A, using the identity cosec²A = 1 + cot²A.

Solution:

LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)

Dividing the numerator and denominator by sin A, we get;

= (cot A – 1 + cosec A)/(cot A + 1 – cosec A)

Using the identity cosec²A = 1 + cot²A ⇒ cosec²A – cot²A = 1,

= [cot A – (cosec²A – cot²A) + cosec A]/ (cot A + 1 – cosec A)

= [(cosec A + cot A) – (cosec A – cot A)(cosec A + cot A)] / (cot A + 1 – cosec A)

= [(cosec A + cot A) (1 – cosec A + cot A)]/ (1 – cosec A + cot A)]

= cosec A + cot A

= RHS

Hence proved.

9. Prove that: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

[Hint: Simplify LHS and RHS separately]

Solution:

LHS = (cosec A – sin A)(sec A – cos A)

= [(1/sin A) – sin A) [(1/cos A) – cos A]

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

Using the identity sin²A + cos²A = 1,

= (cos²A/sin A) (sin²A/cos A)

= cos A sin A….(i)

RHS = 1/(tan A + cot A)

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

= (sin A cos A)/ (sin²A + cos²A)

= (sin A cos A)/1

= sin A cos A….(ii)

From (i) and (ii),

LHS = RHS

i.e. (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

Hence proved.

10. If a sin θ + b cos θ = c, prove that a cosθ – b sinθ = √(a² + b² – c²).

Solution:

Given,

a sin θ + b cos θ = c

Squaring on both sides,

(a sin θ + b cos θ)² = c²

a² sin²θ + b² cos²θ + 2ab sin θ cos θ = c²

Using the identity sin²A + cos²A = 1,

a²(1 – cos²θ) + b²(1 – sin²θ) + 2ab sin θ cos θ = c²

a² – a² cos²θ + b² – b² sin²θ + 2ab sin θ cos θ = c²

a² + b² – c² = a² cos²θ + b² sin²θ – 2ab sin θ cos θ

a² + b² – c² = (a cos θ – b sin θ )²

⇒ a cos θ – b sin θ = √(a² + b² – c²)

Hence proved.

Video Lesson on Trigonometry

Practice Questions on Trigonometry

Solve the following trigonometry problems.

1. Prove that (sin α + cos α) (tan α + cot α) = sec α + cosec α.
2. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
3. If sin θ + cos θ = √3, prove that tan θ + cot θ = 1.
4. Evaluate: 2 tan²45° + cos²30° – sin²60°
5. Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.