NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.3 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3 help students understand the values of trigonometric ratios at specific angles like 0°, 30°, 45°, 60°, and 90°. These solutions are prepared as per the latest CBSE syllabus and provide clear, step-by-step explanations.

Exercise 8.3 mainly focuses on:

NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.3 Introduction to Trigonometry

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]

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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]

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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]

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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]

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Introduction to Trigonometry

Easy

Q.

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

[CBSE - 2025]

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Introduction to Trigonometry

Easy

Q.

​$$\rm { Evaluate :\ } 2 \sqrt{2} \cos 45^{\circ} \sin 30^{\circ}+2 \sqrt{3} \cos 30^{\circ}$$​​

[CBSE - 2024]

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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 - 2025]

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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]

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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]

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Introduction to Trigonometry

Easy

Q.

If $\cos (\alpha+\beta)=0$, then find the value of $\cos \left(\frac{\alpha+\beta}{2}\right)$.

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Introduction to Trigonometry

Medium

Q.

If $\sec \theta - \tan \theta = m$, then find the value of $\sec \theta + \tan \theta$ in terms of $m$. Justify your answer using a relevant trigonometric identity.

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Introduction to Trigonometry

Medium

Q.

If $2 \tan A = 3$, then evaluate the expression $\frac{4 \sin A + 3 \cos A}{4 \sin A - 3 \cos A}$.

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• Finding the exact values of sin θ, cos θ, and tan θ for standard angles

• Using known right-angled triangles (30°–60°–90° and 45°–45°–90°)

• Applying trigonometric ratios in numerical problems

• Understanding how trigonometric values change with angles

This exercise is very important because these standard values are frequently used in later chapters and board exam questions.

The solutions are structured in a simple manner to help students memorize and apply these values confidently in examinations.

✅ Q1. Evaluate:

(i)

$\dfrac{\sin 18^\circ}{\cos 72^\circ}$

cos72∘sin18∘​
(ii)

$\dfrac{\tan 26^\circ}{\cot 64^\circ}$

cot64∘tan26∘​
(iii)

$\cos 48^\circ - \sin 42^\circ$

cos48∘−sin42∘
(iv)

$\cosec 31^\circ - \sec 59^\circ$

cosec31∘−sec59∘

Solution:

(i) $\dfrac{\sin 18^\circ}{\cos 72^\circ}$

cos72∘sin18∘​

We know:

$\sin(90^\circ-\theta)=\cos\theta$

sin(90∘−θ)=cosθ

Here,

$\theta = 72^\circ$

θ=72∘

$\sin(90^\circ-72^\circ)=\sin 18^\circ=\cos 72^\circ$

sin(90∘−72∘)=sin18∘=cos72∘

So,

$$\dfrac{\sin 18^\circ}{\cos 72^\circ}=\dfrac{\cos 72^\circ}{\cos 72^\circ}=1$$

cos72∘sin18∘​=cos72∘cos72∘​=1

Answer: 1

(ii) $\dfrac{\tan 26^\circ}{\cot 64^\circ}$

cot64∘tan26∘​

We know:

$\tan(90^\circ-\theta)=\cot\theta$

tan(90∘−θ)=cotθ

Here,

$\theta = 64^\circ$

θ=64∘

$\tan(90^\circ-64^\circ)=\tan 26^\circ=\cot 64^\circ$

tan(90∘−64∘)=tan26∘=cot64∘

So,

$$\dfrac{\tan 26^\circ}{\cot 64^\circ}=\dfrac{\cot 64^\circ}{\cot 64^\circ}=1$$

cot64∘tan26∘​=cot64∘cot64∘​=1

Answer: 1

(iii) $\cos 48^\circ - \sin 42^\circ$

cos48∘−sin42∘

We know:

$\sin(90^\circ-\theta)=\cos\theta$

sin(90∘−θ)=cosθ

Here,

$\theta = 48^\circ$

θ=48∘

$\sin(90^\circ-48^\circ)=\sin 42^\circ=\cos 48^\circ$

sin(90∘−48∘)=sin42∘=cos48∘

So,

$$\cos 48^\circ - \sin 42^\circ=\cos 48^\circ-\cos 48^\circ=0$$

cos48∘−sin42∘=cos48∘−cos48∘=0

Answer: 0

(iv) $\cosec 31^\circ - \sec 59^\circ$

cosec31∘−sec59∘

We know:

$\sec(90^\circ-\theta)=\cosec\theta$

sec(90∘−θ)=cosecθ

Here,

$\theta = 31^\circ$

θ=31∘

$\sec(90^\circ-31^\circ)=\sec 59^\circ=\cosec 31^\circ$

sec(90∘−31∘)=sec59∘=cosec31∘

So,

$$\cosec 31^\circ - \sec 59^\circ=\cosec 31^\circ-\cosec 31^\circ=0$$

cosec31∘−sec59∘=cosec31∘−cosec31∘=0

Answer: 0

✅ Q2. Show that:

(i)

$\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ = 1$

tan48∘tan23∘tan42∘tan67∘=1
(ii)

$\cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ = 0$

cos38∘cos52∘−sin38∘sin52∘=0

Solution:

(i) $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ = 1$

tan48∘tan23∘tan42∘tan67∘=1

LHS:

$$\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ$$

tan48∘tan23∘tan42∘tan67∘

We know:

$\tan(90^\circ-\theta)=\cot\theta$

tan(90∘−θ)=cotθ

$\tan 48^\circ = \tan(90^\circ-42^\circ)=\cot 42^\circ$

tan48∘=tan(90∘−42∘)=cot42∘

$\tan 23^\circ = \tan(90^\circ-67^\circ)=\cot 67^\circ$

tan23∘=tan(90∘−67∘)=cot67∘

So,

$$= (\cot 42^\circ)(\cot 67^\circ)(\tan 42^\circ)(\tan 67^\circ)$$

=(cot42∘)(cot67∘)(tan42∘)(tan67∘)

$$= (\cot 42^\circ \tan 42^\circ)(\cot 67^\circ \tan 67^\circ)$$

=(cot42∘tan42∘)(cot67∘tan67∘)

$$= (1)(1)=1$$

=(1)(1)=1

Hence proved.

(ii) $\cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ = 0$

cos38∘cos52∘−sin38∘sin52∘=0

LHS:

$$\cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ$$

cos38∘cos52∘−sin38∘sin52∘

We know:

$\sin(90^\circ-\theta)=\cos\theta$

sin(90∘−θ)=cosθ

$\sin 38^\circ = \cos 52^\circ$

sin38∘=cos52∘

$\sin 52^\circ = \cos 38^\circ$

sin52∘=cos38∘

So,

$$= \cos 38^\circ \cos 52^\circ - (\cos 52^\circ)(\cos 38^\circ)=0$$

=cos38∘cos52∘−(cos52∘)(cos38∘)=0

Hence proved.

✅ Q3. If $\tan 2A = \cot(A-18^\circ)$

tan2A=cot(A−18∘), where $2A$

$2A$ is acute, find $A$

$A$.

Solution:

We know:

$\cot \theta = \tan(90^\circ-\theta)$

cotθ=tan(90∘−θ)

So,

$$\cot(A-18^\circ)=\tan(90^\circ-(A-18^\circ))=\tan(108^\circ-A)$$

cot(A−18∘)=tan(90∘−(A−18∘))=tan(108∘−A)

Given:

$$\tan 2A = \tan(108^\circ-A)$$

tan2A=tan(108∘−A)

So,

$$2A = 108^\circ - A$$

2A=108∘−A

$$3A = 108^\circ$$

3A=108∘

$$A = 36^\circ$$

A=36∘

Answer:

$A = 36^\circ$

A=36∘

✅ Q4. If $\tan A = \cot B$

$\tan A=\cot B$, prove that $A + B = 90^\circ$

A+B=90∘.

Solution:

Given:

$\tan A = \cot B$

tanA=cotB

But

$\cot B = \tan(90^\circ - B)$

cotB=tan(90∘−B)

So,

$$\tan A = \tan(90^\circ - B)$$

tanA=tan(90∘−B)

$$A = 90^\circ - B$$

A=90∘−B

$$A + B = 90^\circ$$

A+B=90∘

Hence proved.

✅ Q5. If $\sec 4A = \cosec(A-20^\circ)$

sec4A=cosec(A−20∘), where $4A$

$4A$ is acute, find $A$

$A$.

Solution:

We know:

$\sec \theta = \cosec(90^\circ-\theta)$

secθ=cosec(90∘−θ)

So,

$$\sec 4A = \cosec(90^\circ-4A)$$

sec4A=cosec(90∘−4A)

Given:

$$\cosec(90^\circ-4A)=\cosec(A-20^\circ)$$

cosec(90∘−4A)=cosec(A−20∘)

So,

$$90^\circ - 4A = A - 20^\circ$$

90∘−4A=A−20∘

$$110^\circ = 5A$$

110∘=5A

$$A = 22^\circ$$

A=22∘

Answer:

$A = 22^\circ$

A=22∘

✅ Q6. If $A, B, C$

$A,B,C$ are interior angles of triangle $ABC$

$ABC$, show that

$$\sin\left(\frac{B+C}{2}\right)=\cos\left(\frac{A}{2}\right)$$

sin(2B+C​)=cos(2A​)

Solution:

In a triangle:

$$A + B + C = 180^\circ$$

A+B+C=180∘

$$B + C = 180^\circ - A$$

B+C=180∘−A

Divide by 2:

$$\frac{B+C}{2} = 90^\circ - \frac{A}{2}$$

2B+C​=90∘−2A​

Now take sine both sides:

$$\sin\left(\frac{B+C}{2}\right)=\sin\left(90^\circ-\frac{A}{2}\right)$$

sin(2B+C​)=sin(90∘−2A​)

But

$\sin(90^\circ-\theta)=\cos\theta$

sin(90∘−θ)=cosθ

So,

$$\sin\left(\frac{B+C}{2}\right)=\cos\left(\frac{A}{2}\right)$$

sin(2B+C​)=cos(2A​)

Hence proved.

✅ Q7. Express $\sin 67^\circ + \cos 75^\circ$

sin67∘+cos75∘ in terms of angles between $0^\circ$

0∘ and $45^\circ$

45∘.

Solution:

We know:

$\sin \theta = \cos(90^\circ-\theta)$

sinθ=cos(90∘−θ)

$\cos \theta = \sin(90^\circ-\theta)$

cosθ=sin(90∘−θ)

So,

$$\sin 67^\circ = \cos(90^\circ-67^\circ)=\cos 23^\circ$$

sin67∘=cos(90∘−67∘)=cos23∘

$$\cos 75^\circ = \sin(90^\circ-75^\circ)=\sin 15^\circ$$

cos75∘=sin(90∘−75∘)=sin15∘

Therefore,

$$\sin 67^\circ + \cos 75^\circ = \cos 23^\circ + \sin 15^\circ$$

sin67∘+cos75∘=cos23∘+sin15∘

Answer:

$\cos 23^\circ + \sin 15^\circ$

cos23∘+sin15∘

FAQs: Class 10 Maths Chapter 8 – Exercise 8.3

Q1. What is the focus of Exercise 8.3?
Answer:
Exercise 8.3 focuses on finding and using the exact values of trigonometric ratios for standard angles such as 0°, 30°, 45°, 60°, and 90°.

Q2. Why are standard angle values important?
Answer:
Standard angle values are frequently used in solving trigonometric problems in board exams and higher classes. They form the base of many trigonometric calculations.

Q3. What are some important trigonometric values to remember?
Answer:
Some key values include:

• sin 30° = 1/2

• cos 60° = 1/2

• sin 45° = 1/√2

• tan 45° = 1

Q4. How are these values derived?
Answer:
These values are derived using special right-angled triangles:

• 30°–60°–90° triangle

• 45°–45°–90° triangle

Q5. How do NCERT Solutions help in exam preparation?
Answer:
NCERT Solutions provide clear explanations and help students practice using standard trigonometric values correctly, improving speed and accuracy in exams.