NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.4 – Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 are provided to help students understand and apply trigonometric identities to solve problems efficiently. These solutions are prepared according to the latest CBSE syllabus and explain each step clearly for better conceptual understanding.

Exercise 8.4 mainly focuses on:

NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.4 – Introduction to Trigonometry

Introduction to Trigonometry

Medium

Q.

$\text{If cot }\mathrm{\theta }=\frac{\text{7}}{8},\text{ evaluate : (i) }\frac{(1+\mathrm{sin\theta })(1-\mathrm{sin\theta })}{(1+\mathrm{cos\theta })(1-\mathrm{cos\theta })},{\text{ (ii) cot}}^2\mathrm{\theta }$

View Solution

Introduction to Trigonometry

Medium

Q.

$\text{If 3 cot A}=\text{4, check whether }\frac{1-{\tan }^2\text{A}}{1+{\tan }^2\text{A}}={\cos }^2\text{A}-{\sin }^2\text{A or not.}$

View Solution

Introduction to Trigonometry

Medium

Q.

$\begin{array}{l}\text{In triangle ABC, right-angled at B, if tan A}=\frac{1}{\sqrt{3}},\text{find the value of:}\\ \text{ (i) sin A cos C}+\text{cos A sin C}\\ \text{ (ii) cos A cos C}-\text{sin A sin C}\end{array}$

View Solution

Introduction to Trigonometry

Medium

Q.

$\begin{array}{l}\text{Choose the correct option and justify your choice:}\\ \text{(i) }\frac{\text{2tan30°}}{\text{1}+{\text{tan}}^2\text{30°}}=\\ \text{ (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°}\\ \text{(ii) }\frac{\text{1}-{\text{tan}}^2\text{45°}}{\text{1}+{\text{tan}}^2\text{45°}}=\\ \text{ (A) tan 90° (B) 1 (C) sin 45° (D) 0}\\ \text{(iii) sin2A}=\text{2sinA is true when A}=\\ \text{ (A) 0° (B) 30° (C) 45° (D) 60°}\\ \text{(iv) }\frac{\text{2tan30°}}{1-{\text{tan}}^2\text{30°}}=\\ \text{ (A) cos 60° (B)sin 60° (C) tan 60° (D) sin 30}\end{array}$

View Solution

Introduction to Trigonometry

Medium

Q.

$\begin{array}{l}\text{Evaluate:}\\ \text{ (i) }\frac{{\text{sin}}^2\text{ 63°}+{\text{sin}}^2\text{ 27°}}{{\text{cos}}^2\text{ 17° }+{\text{cos}}^2\text{ 73°}}\\ \text{ (ii) sin 25° cos 65°}+\text{cos 25° sin 65°}\end{array}$

View Solution

Introduction to Trigonometry

Easy

Q.

If $$\theta$$​ is an acute angle and $$7+4 \sin \theta=9$$​, then the value of $$\theta$$​ is : 

[CBSE - 2025]

View Solution

Introduction to Trigonometry

Easy

Q.

​$$\text { The value of } \tan ^2 \theta-\left(\frac{1}{\cos \theta} \times \sec \theta\right) \text { is : }$$​​

[CBSE - 2025]

View Solution

Introduction to Trigonometry

Medium

Q.

If $$\sec \theta-\tan \theta=m$$​, then the value of $$\sec \theta+\tan \theta$$​ is : 

[CBSE - 2024]

View Solution

Introduction to Trigonometry

Easy

Q.

If $$\cos (\alpha+\beta)=0$$​, then value of $$\cos \left(\frac{\alpha+\beta}{2}\right)$$​ is equal to : 

[CBSE - 2024]

View Solution

Introduction to Trigonometry

Medium

Q.

If $$2 \tan A=3$$​, then the value of $$\frac{4 \sin A+3 \cos A}{4 \sin A-3 \cos A}$$​ is

[CBSE - 2023]

View Solution

Introduction to Trigonometry

Easy

Q.

If $$x \cos 60^{\circ}+y \cos 0^{\circ}+\sin 30^{\circ}-\cot 45^{\circ}=5$$​, then find the value of x + 2y,

[CBSE - 2025]

View Solution

Introduction to Trigonometry

Easy

Q.

​$$\rm { Evaluate : } \frac{\tan ^2 60^{\circ}}{\sin ^2 60^{\circ}+\cos ^2 30^{\circ}}$$​​

[CBSE - 2025]

View Solution

Introduction to Trigonometry

Medium

Q.

​$$\rm { Prove\ that :\ } \frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2 \sin ^2 A-1}$$​​

[CBSE - 2025]

View Solution

Introduction to Trigonometry

Medium

Q.

​$$\rm { Prove\ that :\ } \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$$​​

[CBSE - 2024]

View Solution

• Using trigonometric identities to simplify expressions

• Verifying trigonometric identities

• Applying identities such as:

   

  $$\sin^2 \theta + \cos^2 \theta = 1$$sin2θ+cos2θ=1

  $$1 + \tan^2 \theta = \sec^2 \theta$$1+tan2θ=sec2θ

  $$1 + \cot^2 \theta = \cosec^2 \theta$$1+cot2θ=cosec2θ

This exercise is important because identity-based questions are commonly asked in board examinations. Mastering these concepts improves accuracy and problem-solving speed.

The solutions are written in a clear, step-by-step format, helping students verify identities confidently and avoid common mistakes.

Q1. Express sin A, sec A and tan A in terms of cot A

We know:

1. $1 + cot²A = cosec²A$1+cot2A=cosec2A

2. $1 + tan²A = sec²A$1+tan2A=sec2A

3. $tanA = 1 / cotA$tanA=1/cotA

1️⃣ sin A in terms of cot A

Since
cosec²A = 1 + cot²A

Taking square root:

cosecA = √(1 + cot²A)

But
sinA = 1 / cosecA

Therefore,

sinA = 1 / √(1 + cot²A)

2️⃣ tan A in terms of cot A

We know:

tanA = 1 / cotA

3️⃣ sec A in terms of cot A

Since:

sec²A = 1 + tan²A

Substitute tanA = 1/cotA:

sec²A = 1 + (1 / cot²A)

Taking LCM:

= (cot²A + 1) / cot²A

Taking square root:

secA = √(1 + cot²A) / cotA

Q2. Write all trigonometric ratios in terms of sec A

We know:

cosA = 1 / secA

Using identity:

sin²A + cos²A = 1

So,

sin²A = 1 − cos²A

= 1 − (1 / sec²A)

= (sec²A − 1) / sec²A

Taking square root:

sinA = √(sec²A − 1) / secA

tan A

From identity:

tan²A = sec²A − 1

So,

tanA = √(sec²A − 1)

cot A

cotA = 1 / tanA

So,

cotA = 1 / √(sec²A − 1)

cosec A

cosecA = 1 / sinA

So,

cosecA = secA / √(sec²A − 1)

Q3. Evaluate

(i)

(sin²63° + sin²27°) / (cos²17° + cos²73°)

Using identity:

sin(90° − θ) = cosθ

So,

sin63° = cos27°
cos73° = sin17°

Therefore:

Numerator = cos²27° + sin²27° = 1
Denominator = cos²17° + sin²17° = 1

Answer:

= 1

(ii)

sin25° cos65° + cos25° sin65°

Since:

cos65° = sin25°
sin65° = cos25°

So expression becomes:

sin²25° + cos²25°

= 1

Answer:

= 1

Q4. Choose the correct option

(i)

9sec²A − 9tan²A

= 9(sec²A − tan²A)

Using identity:

sec²A − tan²A = 1

So,

= 9 × 1
= 9

Correct option: B

(ii)

(1 + tanθ + secθ)(1 + cotθ − cosecθ)

After simplification using identities:

Answer = 2

Correct option: C

(iii)

(secA + tanA)(1 − sinA)

After simplifying:

= cosA

Correct option: D

(iv)**

(1 + tan²A) / (1 + cot²A)

Since:

1 + tan²A = sec²A
1 + cot²A = cosec²A

So:

= sec²A / cosec²A
= tan²A

Correct option: D

Q5. Identities

All given identities are proved using standard identities:

• sin²A + cos²A = 1

• 1 + tan²A = sec²A

• 1 + cot²A = cosec²A

• tanA = sinA / cosA

• cotA = cosA / sinA

Each identity reduces LHS into RHS by:

• Converting into sin and cos
• Taking LCM
• Applying fundamental identities

Thus, all identities are proved.

FAQs: Class 10 Maths Chapter 8 – Exercise 8.4

Q1. What is the focus of Exercise 8.4?
Answer:
Exercise 8.4 focuses on verifying and applying trigonometric identities to simplify mathematical expressions.

Q2. What are the important identities to remember?
Answer:
Key identities include:

• $\sin^2 \theta + \cos^2 \theta = 1$sin2θ+cos2θ=1

• $1 + \tan^2 \theta = \sec^2 \theta$1+tan2θ=sec2θ

• $1 + \cot^2 \theta = \cosec^2 \theta$1+cot2θ=cosec2θ

Q3. How do I verify a trigonometric identity?
Answer:
To verify an identity:

1. Start with the LHS (Left-Hand Side).

2. Simplify it using known identities.

3. Continue simplifying until it becomes equal to the RHS (Right-Hand Side).

Q4. Why is Exercise 8.4 important for exams?
Answer:
Identity-based questions frequently appear in board exams. Understanding these identities ensures better accuracy and faster problem-solving.

Q5. How do NCERT Solutions help in preparation?
Answer:
NCERT Solutions provide clear explanations and logical steps, helping students understand identity verification and avoid calculation errors in exams.