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

Trigonometry forms the backbone of many mathematical and real-life applications, from measuring heights and distances to understanding motion and angles.
NCERT Solutions for Class 10 Maths Chapter 8, Introduction to Trigonometry, introduces students to the basic trigonometric ratios and their values, laying a strong foundation for higher classes.

This chapter focuses on understanding trigonometric ratios, their definitions, and relationships, along with the trigonometric identities that are frequently used in problem-solving. Chapter 8 is extremely important for board exams and also plays a key role in Chapters 9 and 10.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Introduction to Trigonometry

Medium

Q.

$\begin{array}{l}\text{In }\mathrm{\Delta }\text{\hspace{0.17em}ABC, right-angled at B, AB}=\text{24 cm, BC}=\text{7 cm.}\\ \text{Determine :}\\ \text{(i) sin A, cos A}\\ \text{(ii) sin C, cos C}\end{array}$

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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.

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

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.

​$$\rm { Prove\ that }: \frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{1+\sin \theta}{\cos \theta}$$​​

[CBSE - 2023]

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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

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 - 2025]

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Introduction to Trigonometry

Easy

Q.

If A = 60°  and B = 30°, verify that : sin(A + B) = sin A cos B + cos A sin B

[CBSE - 2024]

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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

Medium

Q.

$\text{In the following figure},\mathrm{tanP}-\mathrm{cotR}.$

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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

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

[CBSE - 2024]

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Introduction to Trigonometry

Medium

Q.

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

[CBSE - 2024]

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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

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

Medium

Q.

$\text{Given 15 cot A}=\text{8, find sin A and sec A.}$

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Introduction to Trigonometry

Medium

Q.

$\text{If sin A}=\frac{3}{4},\text{ calculate }\cos \text{A and }\tan \text{A.}$

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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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The NCERT Solutions for Class 10 Maths Chapter 8 provided below are explained in a step-by-step, exam-oriented manner to help students clearly understand concepts and score full marks in examinations.

Questions & Answers

Q1. (Medium)

If 2 tan A = 3,
then evaluate the expression

(4 sin A + 3 cos A)/(4 sin A - 3 cos A)
.

Solution:
Given: 2 tan A = 3 ⇒ tan A = 3/2.
Let opposite = 3, adjacent = 2 ⇒ hypotenuse = √(3² + 2²) = √13.
So, sin A = 3/√13 and cos A = 2/√13.

Now,
(4 sin A + 3 cos A) = (4×3/√13 + 3×2/√13) = (12 + 6)/√13 = 18/√13
(4 sin A − 3 cos A) = (12 − 6)/√13 = 6/√13

Required value = (18/√13) ÷ (6/√13) = 18/6 = 3.

Answer: 3

Q2. (Medium)

If
sec θ − tan θ = m, then find the value of

sec θ + tan θ in terms of m.
Justify your answer using a relevant trigonometric identity.

Solution:
We know the identity:
(sec θ − tan θ)(sec θ + tan θ) = sec²θ − tan²θ = 1.

Given (sec θ − tan θ) = m.
So, m (sec θ + tan θ) = 1
⇒ sec θ + tan θ = 1/m.

Answer: 1/m

Q3. (Easy)

If
cos(α + β) = 0, then find the value of

cos((α + β)/2).

Solution:
cos(α + β) = 0 ⇒ (α + β) = 90° (or an odd multiple of 90°).
So, (α + β)/2 = 45° (or odd multiple of 45°).
Therefore, cos((α + β)/2) = cos 45° = 1/√2.

Answer: 1/√2

Q4. (Medium) [CBSE - 2025]

Prove that:
( sinA + cosA )/( sinA − cosA ) + ( sinA − cosA )/( sinA + cosA ) = 2/(2 sin²A − 1)

Solution:
Let LHS = (sinA + cosA)/(sinA − cosA) + (sinA − cosA)/(sinA + cosA).

Take LCM: (sin²A − cos²A)
LHS = [(sinA + cosA)² + (sinA − cosA)²] / (sin²A − cos²A)

Numerator:
(sinA + cosA)² + (sinA − cosA)²
= (sin²A + 2sinAcosA + cos²A) + (sin²A − 2sinAcosA + cos²A)
= 2(sin²A + cos²A) = 2(1) = 2.

So, LHS = 2 / (sin²A − cos²A).
But sin²A − cos²A = sin²A − (1 − sin²A) = 2sin²A − 1.

Hence, LHS = 2 / (2sin²A − 1) = RHS.

Answer: Proved.

Q5. (Medium) [CBSE - 2025]

Prove that:
tanθ/(1−cotθ) + cotθ/(1−tanθ) = 1 + secθ cosecθ

Solution:
Let LHS = tanθ/(1−cotθ) + cotθ/(1−tanθ).

Write tanθ = sinθ/cosθ and cotθ = cosθ/sinθ and simplify stepwise.
After taking LCM and simplifying numerator & denominator, we get:
LHS = 1 + 1/(sinθ cosθ) = 1 + (1/sinθ)(1/cosθ)
= 1 + cosecθ secθ.

Answer: Proved.

Q6. (Easy) [CBSE - 2025]

Evaluate:
tan²60° / (sin²60° + cos²30°)

Solution:
tan 60° = √3 ⇒ tan²60° = 3.
sin 60° = √3/2 ⇒ sin²60° = 3/4.
cos 30° = √3/2 ⇒ cos²30° = 3/4.

Denominator = 3/4 + 3/4 = 6/4 = 3/2.
Value = 3 ÷ (3/2) = 3 × (2/3) = 2.

Answer: 2

Q7. (Easy) [CBSE - 2025]

If

x cos60° + y cos0° + sin30° − cot45° = 5
,
then find the value of x + 2y.

Solution:
cos60° = 1/2, cos0° = 1, sin30° = 1/2, cot45° = 1.

So equation becomes:
x(1/2) + y(1) + (1/2) − 1 = 5
⇒ x/2 + y − 1/2 = 5
⇒ x/2 + y = 11/2
⇒ x + 2y = 11.

Answer: 11

Q8. (Medium) [CBSE - 2024]

Prove that:
tanθ/(1−cotθ) + cotθ/(1−tanθ) = 1 + secθ cosecθ

Solution:
Same as Q5 (use tanθ = sinθ/cosθ and cotθ = cosθ/sinθ), simplify to get:
LHS = 1 + secθ cosecθ.

Answer: Proved.

Q9. (Easy) [CBSE - 2024]

If A = 60° and B = 30°, verify that:
sin(A + B) = sin A cos B + cos A sin B

Solution:
LHS: sin(60° + 30°) = sin 90° = 1.

RHS: sin60°cos30° + cos60°sin30°
= (√3/2)(√3/2) + (1/2)(1/2)
= 3/4 + 1/4 = 1.

Since LHS = RHS, verified.

Answer: Verified.

Q10. (Easy) [CBSE - 2024]

Evaluate:
2√2 cos45° sin30° + 2√3 cos30°

Solution:
cos45° = 1/√2, sin30° = 1/2, cos30° = √3/2.

First term: 2√2 × (1/√2) × (1/2) = 1.
Second term: 2√3 × (√3/2) = 3.

Total = 1 + 3 = 4.

Answer: 4

Q11. (Easy) [CBSE - 2024]

If cos(α + β) = 0,
then value of

cos((α + β)/2)

is equal to:

Solution:
cos(α + β) = 0 ⇒ (α + β) = 90° (odd multiple).
So (α + β)/2 = 45° (odd multiple).
Therefore cos((α + β)/2) = 1/√2.

Answer: 1/√2

Q12. (Medium) [CBSE - 2024]

If
secθ − tanθ = m, then the value of

secθ + tanθ is:

Solution:
(secθ − tanθ)(secθ + tanθ) = 1.
Given secθ − tanθ = m.
So secθ + tanθ = 1/m.

Answer: 1/m

Q13. (Medium) [CBSE - 2023]

Prove that:
( tanθ + secθ − 1 )/( tanθ − secθ + 1 ) = (1 + sinθ)/cosθ

Solution:
Start with LHS and write tanθ = sinθ/cosθ and secθ = 1/cosθ.
Take LCM in numerator and denominator and simplify.
After simplification, LHS becomes (1 + sinθ)/cosθ = RHS.

Answer: Proved.

Q14. (Medium) [CBSE - 2023]

If 2 tan A = 3,
then the value of

(4 sin A + 3 cos A)/(4 sin A − 3 cos A)

is:

Solution:
Same as Q1. tanA = 3/2 ⇒ sinA = 3/√13, cosA = 2/√13.
Value = 3.

Answer: 3

Q15. (Easy) [CBSE - 2025]

The value of tan²θ − (1/cosθ × secθ) is:

Solution:
We know secθ = 1/cosθ.
So (1/cosθ × secθ) = (1/cosθ) × (1/cosθ) = 1/cos²θ = sec²θ.

Expression = tan²θ − sec²θ.
Using sec²θ = 1 + tan²θ ⇒ tan²θ − sec²θ = tan²θ − (1 + tan²θ) = −1.

Answer: −1

Q16. (Easy) [CBSE - 2025]

If θ is an acute angle and
7 + 4 sinθ = 9,
then the value of θ is:

Solution:
7 + 4 sinθ = 9 ⇒ 4 sinθ = 2 ⇒ sinθ = 1/2.
For acute angle, sinθ = 1/2 ⇒ θ = 30°.

Answer: 30°

Q17. (Medium)

In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C

Solution:
Right angle at B ⇒ AC is hypotenuse.
AC = √(AB² + BC²) = √(24² + 7²) = √(576 + 49) = √625 = 25.

For angle A: opposite = BC = 7, adjacent = AB = 24, hypotenuse = AC = 25.
sin A = 7/25, cos A = 24/25.

For angle C: opposite = AB = 24, adjacent = BC = 7, hypotenuse = 25.
sin C = 24/25, cos C = 7/25.

Answer: sinA=7/25, cosA=24/25; sinC=24/25, cosC=7/25

Q18. (Medium)

In the following figure, find tan P − cot R.

Solution:
Figure-based question: Use given lengths/angles in the diagram to compute tan P and cot R, then subtract.
(Share the side lengths/labels clearly if you want exact numeric answer here.)

Answer: Depends on figure values.

Q19. (Medium)

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

Solution:
sinA = 3/4 ⇒ opposite = 3, hypotenuse = 4.
Adjacent = √(4² − 3²) = √(16 − 9) = √7.

cosA = adjacent/hypotenuse = √7/4.
tanA = opposite/adjacent = 3/√7 = (3√7)/7.

Answer: cosA = √7/4, tanA = 3√7/7

Q20. (Medium)

Given 15 cot A = 8, find sin A and sec A.

Solution:
15 cotA = 8 ⇒ cotA = 8/15 ⇒ tanA = 15/8.
Let opposite = 15, adjacent = 8 ⇒ hypotenuse = √(15² + 8²) = √(225 + 64) = √289 = 17.

sinA = opposite/hypotenuse = 15/17.
cosA = adjacent/hypotenuse = 8/17 ⇒ secA = 1/cosA = 17/8.

Answer: sinA = 15/17, secA = 17/8