Introduction to Trigonometry: Important Questions and Answers for Class 10

Important Questions on Introduction to Trigonometry Class 10 are available here to help students prepare effectively for CBSE board exams. These questions are designed according to the latest CBSE syllabus and exam pattern. Introduction to Trigonometry is one of the most important chapters in Class 10 Maths as it forms the foundation for advanced mathematical concepts in higher classes. Practising these important questions with detailed solutions will help students strengthen their concepts, improve accuracy, and score higher marks in examinations.

Table of Contents

• Key Concepts of Introduction to Trigonometry for Class 10
• Important Questions on Applications of Trigonometry for Class 10

 

Key Concepts of Introduction to Trigonometry for Class 10

Trigonometric Identities:

 

Important Questions on Applications of Trigonometry for Class 10

 

Q1: In triangle ABC, right-angled at B, if AB = 5 cm and BC = 12 cm, find all six trigonometric ratios of angle A.

Solution: Use Pythagoras to find the hypotenuse AC:

AC² = AB² + BC² = 5² + 12² = 25 + 144 = 169 

⇒ AC = 13 cm

For angle A: Opposite = BC = 12, Adjacent = AB = 5, Hypotenuse = AC = 13.

Write all six ratios:

sin A = 12/13, cos A = 5/13, tan A = 12/5

cosec A = 13/12, sec A = 13/5, cot A = 5/12

Q2: If sin A = 3/4, find cos A and tan A.

Solution: Use the identity sin² A + cos² A = 1:

cos²A = 1 – sin²A A = 1 – 9/16 = 7/16 

⇒ cos A = √7/4

Find tan A:

tan A = sin A / cos A = (3/4) ÷ (√7/4) = 3/√7

Therefore, cos A = √7/4  and tan A = 3/√7 (rationalised: 3√7/7)

Q3: Evaluate: 2 tan² 45° + cos² 30° − sin² 60°

Solution: Substitute the known values: tan 45° = 1, cos 30° = √3/2, sin 60° = √3/2

= 2(1)² + (√3/2)² − (√3/2)²

= 2(1) + 3/4 − 3/4 = 2 + 0 = 2

2tan²45° + cos²30° − sin²60° = 2

Q4: Find the value of (sin 30° + cos 60°) − (sin 60° + cos 30°)

Solution: Substitute values: sin 30° = 1/2, cos 60° = 1/2, sin 60° = √3/2, cos 30° = √3/2

= (1/2 + 1/2) − (√3/2 + √3/2)

= 1 − √3

Answer = 1 − √3 ≈ −0.732

Q5: Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution: Use the complementary rule: sin θ = cos (90° − θ)

sin 67° = cos(90° − 67°) = cos 23°

Use: cos θ = sin(90° − θ)

cos 75° = sin(90° − 75°) = sin 15°

Both 23° and 15° lie between 0° and 45°, so:

sin 67° + cos 75° = cos 23° + sin 15°

Answer: cos 23° + sin 15°

Q6: Evaluate: (sin 18° / cos 72°) + (√3 tan 10° · tan 80°)

Solution: 18° and 72° are complementary (18 + 72 = 90°). 

So cos 72° = sin 18°.

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

Similarly, 10° and 80° are complementary. So tan 80° = cot 10°:

tan 10° · tan 80° = tan 10° · cot 10° = tan 10° × (1/tan 10°) = 1

= 1 + √3 × 1 = 1 + √3

Answer = 1 + √3

Q7: Prove that (1 − cos A) / sin A = sin A / (1 + cos A)

Solution: Multiply numerator and denominator by (1 + cos A):

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

The numerator is a difference of squares: (1 − cos A)(1 + cos A) = 1 − cos²A

= (1 − cos²A) / [sin A (1 + cos A)]

Use identity: 1 − cos²A = sin²A

= sin²A / [sin A (1 + cos A)]

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

LHS = RHS. Hence proved.

Q8: Prove that: (sin θ − cos θ)/(sin θ + cos θ) = (tan θ − 1)/(tan θ + 1)

Solution: tan θ = sin θ/cos θ.

RHS = (tan θ − 1)/(tan θ + 1) = (sin θ/cos θ − 1)/(sin θ/cos θ + 1)

Multiply the numerator and denominator by cos θ:

= (sin θ − cos θ)/(sin θ + cos θ) = LHS

RHS = LHS. Hence proved.

Q9: Prove that (tan A + sin A)/(tan A − sin A) = (sec A + 1)/(sec A − 1)

Solution: tan A = sin A/cos A:

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

Factor sin A from the numerator and denominator:

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

Cancel sin A (since sin A ≠ 0 for acute angles):

= (1/cos A + 1) / (1/cos A − 1)

Since 1/cos A = sec A:

= (sec A + 1) / (sec A − 1) = RHS

LHS = RHS. Hence proved.

 

Q10: Prove that (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A)

Solution: Expand LHS.

(cosec A − sin A) = (1 − sin²A)/sin A = cos²A/sin A

(sec A − cos A) = (1 − cos²A)/cos A = sin²A/cos A

Multiply them:

LHS = (cos²A/sin A) × (sin²A/cos A) = sin A · cos A

Now simplify RHS: tan A + cot A = sin A/cos A + cos A/sin A = (sin²A + cos²A)/(sin A cos A) = 1/(sin A cos A)

So RHS = 1 / [1/(sin A cos A)] = sin A cos A = LHS

LHS = RHS. Hence proved.

Q11: If 1 + sin²θ = 3 sin θ cos θ, prove that tan θ = 1 or tan θ = 1/2.

Solution: Divide both sides by sin² θ:

(1/sin²θ) + 1 = 3(cos θ/sin θ)

cosec²θ + 1 = 3 cot θ

Use the identity cosec²θ = 1 + cot²θ:

(1 + cot²θ) + 1 = 3 cot θ 

cot²θ − 3 cot θ + 2 = 0

Factorise the quadratic:

(cot θ − 1)(cot θ − 2) = 0

cot θ = 1 or cot θ = 2

Since tan θ = 1/cot θ:

tan θ = 1 or tan θ = 1/2

Hence proved: tan θ = 1 or tan θ = 1/2.

Q12: If x sin³θ + y cos³θ = sin θ cos θ and x sin θ = y cos θ, prove that x² + y² = 1.

Solution: From the second equation: x sin θ = y cos θ 

⇒ x/y = cos θ/sin θ

⇒ x = y cos θ/sin θ

Substitute into the first equation:

x sin³ θ + y cos³ θ = sin θ cos θ

(y cos θ/sin θ) sin³ θ + y cos³ θ = sin θ cos θ

y cos θ sin² θ + y cos³ θ = sin θ cos θ

Factor the left side:

y cos θ (sin² θ + cos² θ) = sin θ cos θ

⇒ y cos θ (1) = sin θ cos θ

⇒ y = sin θ

From x sin θ = y cos θ: x sin θ = sin θ cos θ

⇒ x = cos θ

x² + y² = cos²θ + sin²θ = 1

x² + y² = 1    

Hence proved.

Q13: In triangle ABC, right-angled at B, if tan A = 1/√3, find: 

(a) sin A cos C + cos A sin C   (b) cos A cos C − sin A sin C

Solution: tan A = 1/√3 means angle A = 30°. Since ∠B = 90°, angle C = 60°.

Therefore, sin A = sin 30° = 1/2, cos A = cos 30° = √3/2, sin C = sin 60° = √3/2, cos C = cos 60° = 1/2

(a) sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2) (√3/2) = 1/4 + 3/4 = 1

(b) cos A cos C − sin A sin C = (√3/2)(1/2) − (1/2)(√3/2) = √3/4 − √3/4 = 0

(a) = 1 and (b) = 0

Q14: If tan 2A = cot(A − 18°), where 2A is an acute angle, find the value of A.

Solution: tan 2A = cot(A−18°) 

⇒ cot(90°−2A) = cot(A−18°) 

⇒ 90°−2A = A−18° 

⇒ 3A = 108° 

⇒ A = 36°

Q15: If cosec θ − sin θ = l and sec θ − cos θ = m, prove that l²m²(l² + m² + 3) = 1.

Solution: Given cosec θ − sin θ = l and sec θ − cos θ = m, 

l = cosec θ − sin θ  = 1/sin θ - sin θ = (1-sin²θ)/sinθ

Similarly m =   (1-cos²θ)/cosθ

l = cos²θ/sin θ, and m = sin²θ/cos θ. 

Then l²m² = cos²θ sin²θ and l² + m² = cos⁴θ/sin²θ + sin⁴θ/cos²θ.

Using the identity:

a³+b³=(a+b)(a²−ab+b²)

Take a = sin²θ and b = cos²θ

sin⁶θ + cos⁶θ = (sin²θ + cos²θ)(sin⁴θ - sin²θcos²θ + cos⁴θ) = sin⁴θ - sin²θcos²θ + cos⁴θ

Also, sin⁴θ + cos⁴θ = (sin²θ + cos²θ) - 2sin²θcos²θ = 1 - 2sin²θcos²θ.

Hence, sin⁶θ + cos⁶θ = 1 - 3sin²θcos²θ.

Therefore, l² + m² = (1 - 3sin²θcos²θ)/sin²θcos²θ = (1/sin²θcos²θ) - 3

l² + m² + 3 = 1/sin²θcos²θ

Multiply both sides by l²m² = cos²θ sin²θ. 

l² + m²(l² + m² + 3) = 1

Hence proved.

Learn more: Trigonometric Identities: Formulas, Functions & Examples