NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry help students understand the basic relationship between the angles and sides of a right-angled triangle. Trigonometry is an important part of mathematics and is used in many real-life fields, such as engineering, architecture, astronomy, navigation, construction, and map-making. This chapter builds the foundation required for learning advanced trigonometry in higher classes.

In this chapter, students learn about the six trigonometric ratios: sine, cosine, tangent, cosecant, secant, and cotangent. These ratios are commonly written as sin, cos, tan, cosec, sec, and cot. Each ratio compares two sides of a right-angled triangle. The three sides are called the hypotenuse, perpendicular, and base. Students must identify these sides correctly according to the angle given in the question.

The chapter also explains trigonometric ratios of specific angles, including 0°, 30°, 45°, 60°, and 90°. Students learn how to find and remember the values of different ratios at these angles. A trigonometric table is useful for solving many questions quickly. The chapter also introduces relationships between complementary angles. For example, the sine of an angle is equal to the cosine of its complementary angle.

Another important part of Chapter 8 is trigonometric identities. An identity is an equation that remains true for all allowed values of an angle. Students learn the main identity involving sine and cosine and use it to prove other expressions. These questions require careful substitution, simplification, and knowledge of basic algebra.

The NCERT Solutions for Class 10 Maths Chapter 8 provide clear, step-by-step answers to all textbook exercises. Each solution explains the formula used, the values substituted, and the calculation needed to reach the final answer. The simple method helps students understand why a particular ratio or identity is applied.

These solutions are useful for completing homework, preparing class notes, revising formulas, and studying for the CBSE Class 10 board examination. Students should first attempt each question independently and then check the solution. Regular practice will help them remember important values, avoid sign errors, and improve calculation speed.

Download Class 10 Maths Chapter 8 Introduction to Trigonometry NCERT Solutions PDF

Download the Class 10 Maths Chapter 8 Introduction to Trigonometry NCERT Solutions PDF given below to study the chapter in a simple and organised way. It contains clear, step-by-step answers to the textbook questions on trigonometric ratios, standard angle values, complementary angles, and important trigonometric identities. These solutions can help students understand the correct method, complete homework, revise key concepts, and prepare for school tests and the CBSE Class 10 board examination.

Access NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Exercise 8.1

Question 1: In ΔABC, right-angled at B, AB = 24 cm and BC = 7 cm. Determine:

(i) sin A and cos A

(ii) sin C and cos C

Solution:

Since the triangle is right-angled at B, AC is the hypotenuse.

Using the Pythagoras theorem:

AC² = AB² + BC²

AC² = 24² + 7² = 576 + 49 = 625

AC = √625 = 25 cm

(i) Ratios for angle A:

sin A = BC/AC = 7/25

cos A = AB/AC = 24/25

(ii) Ratios for angle C:

sin C = AB/AC = 24/25

cos C = BC/AC = 7/25

Final answer: sin A = 7/25, cos A = 24/25, sin C = 24/25 and cos C = 7/25.

Question 2: In a right-angled triangle PQR, PQ = 12 cm and PR = 13 cm. Find tan P − cot R.

Solution:

PR is the hypotenuse. Let QR be the third side.

PR² = PQ² + QR²

13² = 12² + QR²

169 = 144 + QR²

QR² = 25, so QR = 5 cm

tan P = QR/PQ = 5/12

For angle R, cot R = QR/PQ = 5/12

tan P − cot R = 5/12 − 5/12 = 0

Final answer: 0

Question 3: If sin A = 3/4, calculate cos A and tan A.

Solution:

sin A = (Opposite side)/(Hypotenuse) = 3/4

Take the opposite side as 3k and the hypotenuse as 4k, where k is a positive number.

Using the Pythagoras theorem:

(Adjacent side)² = (4k)² − (3k)²

(Adjacent side)² = 16k² − 9k² = 7k²

Adjacent side = √7k

cos A = (√7k)/(4k) = √7/4

tan A = (3k)/(√7k) = 3/√7 = 3√7/7

Final answer: cos A = √7/4 and tan A = 3√7/7.

Question 4: Given that 15 cot A = 8, find sin A and sec A.

Solution:

15 cot A = 8

cot A = 8/15

Since cot A = (Adjacent side)/(Opposite side), take:

Adjacent side = 8k and opposite side = 15k

Hypotenuse² = (8k)² + (15k)²

Hypotenuse² = 64k² + 225k² = 289k²

Hypotenuse = 17k

sin A = (15k)/(17k) = 15/17

sec A = (17k)/(8k) = 17/8

Final answer: sin A = 15/17 and sec A = 17/8.

Question 5: Given that sec θ = 13/12, calculate all the other trigonometric ratios.

Solution:

sec θ = (Hypotenuse)/(Adjacent side) = 13/12

Take the hypotenuse as 13k and the adjacent side as 12k.

Opposite side² = (13k)² − (12k)²

Opposite side² = 169k² − 144k² = 25k²

Opposite side = 5k

Question 6: If A and B are acute angles such that cos A = cos B, show that A = B.

Solution:

For acute angles, the value of cosine decreases as the angle increases.

Suppose A were greater than B. Then cos A would be less than cos B, which would contradict the given condition cos A = cos B.

Similarly, A cannot be less than B because that would make cos A greater than cos B.

Therefore, the only possible result is A = B.

Hence proved.

Question 7: If cot θ = 7/8, evaluate:

(i) ((1 + sin θ)(1 − sin θ))/((1 + cos θ)(1 − cos θ))

(ii) cot² θ

Solution:

(i)

((1 + sin θ)(1 − sin θ))/((1 + cos θ)(1 − cos θ))

= (1 − sin² θ)/(1 − cos² θ)

= cos² θ/sin² θ

= cot² θ

= (7/8)² = 49/64

(ii)

cot² θ = (7/8)² = 49/64

Final answer: (i) 49/64 and (ii) 49/64.

Question 8: If 3 cot A = 4, check whether (1 − tan² A)/(1 + tan² A) = cos² A − sin² A.

Solution:

3 cot A = 4, so cot A = 4/3 and tan A = 3/4.

Take the adjacent side as 4k and the opposite side as 3k. The hypotenuse is 5k.

Left-hand side:

(1 − tan² A)/(1 + tan² A)

= (1 − (3/4)²)/(1 + (3/4)²)

= (1 − 9/16)/(1 + 9/16)

= (7/16)/(25/16) = 7/25

Right-hand side:

cos A = 4/5 and sin A = 3/5

cos² A − sin² A = (4/5)² − (3/5)²

= 16/25 − 9/25 = 7/25

Both sides are equal to 7/25.

Therefore, the given equality is true.

Question 9: In ΔABC, right-angled at B, tan A = 1/√3. Find:

(i) sin A cos C + cos A sin C

(ii) cos A cos C − sin A sin C

Solution:

tan A = BC/AB = 1/√3

Take BC = k and AB = √3k.

AC² = k² + (√3k)² = 4k²

AC = 2k

(i)

sin A cos C + cos A sin C

= (1/2)(1/2) + (√3/2)(√3/2)

= 1/4 + 3/4 = 1

(ii)

cos A cos C − sin A sin C

= (√3/2)(1/2) − (1/2)(√3/2)

= √3/4 − √3/4 = 0

Final answer: (i) 1 and (ii) 0.

Question 10: In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.

Solution:

Let QR = x cm. Then PR = 25 − x cm.

Since PR is the hypotenuse:

PR² = PQ² + QR²

(25 − x)² = 5² + x²

625 − 50x + x² = 25 + x²

625 − 50x = 25

50x = 600

x = 12

Therefore, QR = 12 cm and PR = 13 cm.

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

Final answer: sin P = 12/13, cos P = 5/13 and tan P = 12/5.

Question 11: State whether each statement is true or false. Give a reason.

Exercise 8.2

Standard Trigonometric Values

Question 1: Evaluate the following expressions.

(i) sin 60° cos 30° + sin 30° cos 60°

= (√3/2)(√3/2) + (1/2)(1/2)

= 3/4 + 1/4 = 1

Answer: 1

(ii) 2 tan² 45° + cos² 30° − sin² 60°

= 2(1)² + (√3/2)² − (√3/2)²

= 2 + 3/4 − 3/4 = 2

Answer: 2

(iii) (cos 45°)/(sec 30° + cosec 30°)

= (√2/2)/((2/√3) + 2)

= (√2/2)/(2(√3 + 1)/√3)

= √6/(4(√3 + 1))

After rationalising the denominator:

= (3√2 − √6)/8

Answer: (3√2 − √6)/8

(iv) (sin 30° + tan 45° − cosec 60°)/(sec 30° + cos 60° + cot 45°)

= (1/2 + 1 − 2/√3)/(2/√3 + 1/2 + 1)

= ((3√3 − 4)/(2√3))/((4 + 3√3)/(2√3))

= (3√3 − 4)/(4 + 3√3)

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

= (3√3 − 4)²/((3√3)² − 4²)

= (43 − 24√3)/11

Answer: (43 − 24√3)/11

(v) (5 cos² 60° + 4 sec² 30° − tan² 45°)/(sin² 30° + cos² 30°)

Numerator = 5(1/2)² + 4(2/√3)² − 1²

= 5/4 + 16/3 − 1

= 1/4 + 16/3 = 67/12

Denominator = sin² 30° + cos² 30° = 1

Therefore, the value is (67/12)/1 = 67/12.

Answer: 67/12

Question 2: Choose the correct option and justify the answer.

(i) (2 tan 30°)/(1 + tan² 30°)

= (2/√3)/(1 + 1/3)

= (2/√3)/(4/3) = √3/2

Since sin 60° = √3/2, the correct option is (a) sin 60°.

(ii) (1 − tan² 45°)/(1 + tan² 45°)

= (1 − 1)/(1 + 1) = 0/2 = 0

The correct option is (d) 0.

(iii) sin 2A = 2 sin A is true when A equals:

For A = 0°:

sin(2 × 0°) = sin 0° = 0

2 sin 0° = 0

Both sides are equal. The correct option is (a) 0°.

(iv) (2 tan 30°)/(1 − tan² 30°)

= (2/√3)/(1 − 1/3)

= (2/√3)/(2/3) = √3

Since tan 60° = √3, the correct option is (c) tan 60°.

Question 3: 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°

Therefore, A + B = 60°. ...(1)

tan(A − B) = 1/√3 = tan 30°

Therefore, A − B = 30°. ...(2)

Adding equations (1) and (2):

2A = 90°

A = 45°

Substituting A = 45° in A + B = 60°:

45° + B = 60°

B = 15°

Final answer: A = 45° and B = 15°.

Question 4: State whether each statement is true or false. Give a reason.

Exercise 8.3

Question 1: Express sin A, sec A and tan A in terms of cot A.

Solution:

Use the identity:

cosec² A = 1 + cot² A

Since cosec A = 1/(sin A):

1/(sin² A) = 1 + cot² A

sin² A = 1/(1 + cot² A)

Because A is acute, sin A is positive.

sin A = 1/√(1 + cot² A)

Also:

tan A = 1/(cot A)

Using sec² A = 1 + tan² A:

sec² A = 1 + 1/(cot² A)

sec² A = (1 + cot² A)/(cot² A)

Since A is acute:

sec A = √(1 + cot² A)/(cot A)

Final answer:

• sin A = 1/√(1 + cot² A)
• sec A = √(1 + cot² A)/(cot A)
• tan A = 1/(cot A)

Question 2: Write all the other trigonometric ratios of A in terms of sec A.

Solution:

cos A = 1/(sec A)

Using sec² A = 1 + tan² A:

tan² A = sec² A − 1

tan A = √(sec² A − 1)

Now:

sin A = tan A × cos A

sin A = √(sec² A − 1)/(sec A)

cot A = 1/(tan A)

cot A = 1/√(sec² A − 1)

cosec A = 1/(sin A)

cosec A = (sec A)/√(sec² A − 1)

Question 3: Evaluate the following.

(i) (sin² 63° + sin² 27°)/(cos² 17° + cos² 73°)

Since 63° = 90° − 27°:

sin 63° = cos 27°

Therefore, the numerator is:

cos² 27° + sin² 27° = 1

Also, 17° = 90° − 73°, so cos 17° = sin 73°.

Therefore, the denominator is:

sin² 73° + cos² 73° = 1

Hence, the value is 1/1 = 1.

Answer: 1

(ii) sin 25° cos 65° + cos 25° sin 65°

Since cos 65° = sin 25° and sin 65° = cos 25°:

sin 25° cos 65° + cos 25° sin 65°

= sin² 25° + cos² 25°

= 1

Answer: 1

Question 4: Choose the correct option and justify the answer.

(i) 9 sec² A − 9 tan² A

= 9(sec² A − tan² A)

= 9(1) = 9

The correct option is (b) 9.

(ii) (1 + tan θ + sec θ)(1 + cot θ − cosec θ)

Write all ratios in terms of sin θ and cos θ:

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

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

= ((sin θ + cos θ)² − 1)/(sin θ cos θ)

= (sin² θ + cos² θ + 2 sin θ cos θ − 1)/(sin θ cos θ)

= (2 sin θ cos θ)/(sin θ cos θ) = 2

The correct option is (c) 2.

(iii) (sec A + tan A)(1 − sin A)

= ((1 + sin A)/(cos A))(1 − sin A)

= ((1 + sin A)(1 − sin A))/(cos A)

= (1 − sin² A)/(cos A)

= cos² A/(cos A) = cos A

The correct option is (d) cos A.

(iv) (1 + tan² A)/(1 + cot² A)

= sec² A/cosec² A

= (1/cos² A)/(1/sin² A)

= sin² A/cos² A

= tan² A

The correct option is (d) tan² A.

Question 5: Prove the following identities. The angles are acute, and every expression is assumed to be defined.

(i) (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ)

Left-hand side:

(cosec θ − cot θ)²

= (1/(sin θ) − cos θ/(sin θ))²

= ((1 − cos θ)/(sin θ))²

= (1 − cos θ)²/sin² θ

= (1 − cos θ)²/(1 − cos² θ)

= (1 − cos θ)²/((1 − cos θ)(1 + cos θ))

= (1 − cos θ)/(1 + cos θ)

This is the right-hand side. Hence proved.

(ii) (cos A)/(1 + sin A) + (1 + sin A)/(cos A) = 2 sec A

Left-hand side:

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

= (cos² A + (1 + sin A)²)/((1 + sin A)cos A)

= (cos² A + 1 + 2 sin A + sin² A)/((1 + sin A)cos A)

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

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

= 2/(cos A) = 2 sec A

This is the right-hand side. Hence proved.

(iii) (tan θ)/(1 − cot θ) + (cot θ)/(1 − tan θ) = 1 + sec θ cosec θ

Left-hand side:

(tan θ)/(1 − cot θ) + (cot θ)/(1 − tan θ)

= (sin² θ)/(cos θ(sin θ − cos θ)) − (cos² θ)/(sin θ(sin θ − cos θ))

= (sin³ θ − cos³ θ)/(sin θ cos θ(sin θ − cos θ))

= ((sin θ − cos θ)(sin² θ + sin θ cos θ + cos² θ))/(sin θ cos θ(sin θ − cos θ))

= (1 + sin θ cos θ)/(sin θ cos θ)

= 1 + 1/(sin θ cos θ)

= 1 + sec θ cosec θ

This is the right-hand side. Hence proved.

(iv) (1 + sec A)/(sec A) = sin² A/(1 − cos A)

Left-hand side:

(1 + sec A)/(sec A) = 1/(sec A) + 1 = cos A + 1

Right-hand side:

sin² A/(1 − cos A)

= (1 − cos² A)/(1 − cos A)

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

= 1 + cos A

Both sides are equal. Hence proved.

(v) (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A

Left-hand side:

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

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

The numerator becomes:

(1 + cos A)² − sin² A

= 1 + 2 cos A + cos² A − sin² A

= 2 cos² A + 2 cos A

= 2 cos A(1 + cos A)

The denominator becomes:

(cos A + sin A)² − 1

= cos² A + sin² A + 2 sin A cos A − 1

= 2 sin A cos A

Therefore:

Left-hand side = (1 + cos A)/(sin A)

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

= cosec A + cot A

This is the right-hand side. Hence proved.

(vi) √((1 + sin A)/(1 − sin A)) = sec A + tan A

Left-hand side:

√((1 + sin A)/(1 − sin A))

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

= √((1 + sin A)²/(1 − sin² A))

= √((1 + sin A)²/cos² A)

Since A is acute, cos A is positive.

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

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

= sec A + tan A

This is the right-hand side. Hence proved.

(vii) (sin θ − 2 sin³ θ)/(2 cos³ θ − cos θ) = tan θ

Left-hand side:

(sin θ(1 − 2 sin² θ))/(cos θ(2 cos² θ − 1))

Since cos² θ = 1 − sin² θ:

2 cos² θ − 1 = 2(1 − sin² θ) − 1

= 1 − 2 sin² θ

Therefore:

Left-hand side = (sin θ(1 − 2 sin² θ))/(cos θ(1 − 2 sin² θ))

= sin θ/cos θ = tan θ

This is the right-hand side. Hence proved.

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

Left-hand side:

(sin A + cosec A)² + (cos A + sec A)²

= sin² A + 2 sin A cosec A + cosec² A + cos² A + 2 cos A sec A + sec² A

= sin² A + cos² A + 2 + 2 + cosec² A + sec² A

= 1 + 4 + (1 + cot² A) + (1 + tan² A)

= 7 + tan² A + cot² A

This is the right-hand side. Hence proved.

(ix) (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A)

Left-hand side:

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

= ((1 − sin² A)/(sin A))((1 − cos² A)/(cos A))

= (cos² A/(sin A))(sin² A/(cos A))

= sin A cos A

Right-hand side:

1/(tan A + cot A)

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

= 1/((sin² A + cos² A)/(sin A cos A))

= sin A cos A

Both sides are equal. Hence proved.

(x) (1 + tan² A)/(1 + cot² A) = ((1 − tan A)/(1 − cot A))²

Left-hand side:

(1 + tan² A)/(1 + cot² A)

= sec² A/cosec² A

= sin² A/cos² A

= tan² A

Right-hand side:

Let tan A = t. Then cot A = 1/t.

((1 − tan A)/(1 − cot A))²

= ((1 − t)/(1 − 1/t))²

= ((1 − t)/((t − 1)/t))²

= (−t)²

= t² = tan² A

Both sides are equal. Hence proved.