Write all the other trigonometric ratios of \(\angle A\) in terms of \(\sec A\).

Concept used. We use \(\cos A=\tfrac{1}{\sec A}\), the identity \(\sin^2 A=1-\cos^2 A\), and the quotient and reciprocal relations to write \(\sin A\), \(\cos A\), \(\tan A\), \(\cot A\) and \(\csc A\) using only \(\sec A\). For an acute angle all roots are positive. \(\cos A\):

Choose the correct option. Justify your choice: (i) \(9\sec^2 A-9\tan^2 A=\) (A) \(1\) (B) \(9\) (C) \(8\) (D) \(0\) (ii) \((1+\tan\theta+

Concept used. Each part simplifies through the identities \(\sec^2 A-\tan^2 A=1\), \(\sec^2 A=1+\tan^2 A\), \(\csc^2 A=1+\cot^2 A\), and the definitions of the ratios in terms of \(\sin\) and \(\cos\). (i) Factor out \(9\) and use \(\sec^2 A-\tan^2 A=1\): \[ 9\sec^2 A-9\tan^2 A=9

NCERT Solutions Class 10 Maths Chapter 8 Trigonometry Exercise 8.3

Q: Express the trigonometric ratios \(\sin A\), \(\sec A\) and \(\tan A\) in terms of \(\cot A\).

Concept used. We use the identities \(\csc^2 A=1+\cot^2 A\) and \(\sec^2 A=1+\tan^2 A\), together with the reciprocal relation \(\tan A=\tfrac{1}{\cot A}\). Each target ratio is rewritten so that only \(\cot A\) appears. For an acute angle every ratio is positive, so each square

Q: Prove the following identities, where the angles involved are acute angles for which the expressions are defined: (i) \((\csc\theta-\cot\theta)^2=\dfrac{1-\cos\theta}{1+\cos\t

Concept used. Every proof rests on the three identities \(\sin^2\theta+\cos^2\theta=1\), \(\sec^2\theta-\tan^2\theta=1\) and \(\csc^2\theta-\cot^2\theta=1\), plus the definitions \(\tan=\tfrac{\sin}{\cos}\), \(\cot=\tfrac{\cos}{\sin}\), \(\sec=\tfrac{1}{\cos}\), \(\csc=\tfrac{1}{

The NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3 cover all 4 questions on trigonometric identities, according to the latest 2026-27 CBSE syllabus. Each answer is solved step by step exactly the way the board expects.

Questions covered: 4 in total, mixing ratio-rewriting, all-ratios-in-terms-of-one, MCQ simplification, and ten identity proofs.
Core skill: using the three Pythagorean identities to simplify or prove expressions involving trigonometric ratios.
Board value: Exercise 8.3 covers identity proofs and simplification, which appear in 3 to 5 mark slots in the CBSE Class 10 board paper.

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

Solved by Collegedunia: Every Exercise 8.3 question below is solved by subject experts, checked against the official 2026-27 NCERT textbook, and written with full working so each step earns its marks in the CBSE Class 10 paper.

Contents

What Exercise 8.3 Covers for Class 10
How to Solve Exercise 8.3 Question by Question
Three Pythagorean Identities Used in Exercise 8.3
Proof Techniques Students Need for Exercise 8.3
Exercise 8.3 Marks and CBSE Board Trends
All Exercises of Class 10 Maths Chapter 8
More Chapter 8 Resources and Related Pages
All NCERT Solutions for Exercise 8.3 with Step-by-Step Solutions
Exercise 8.3 Trigonometry FAQs

What Exercise 8.3 of Introduction to Trigonometry Covers for Class 10

Exercise 8.3 is the final exercise of Chapter 8 and the hardest one. It moves beyond finding ratios and asks students to prove trigonometric identities and rewrite ratios in terms of a single given ratio. All four questions rest on the three Pythagorean identities at the heart of the chapter.

Question 1: express sin A, sec A and tan A in terms of cot A using the reciprocal square identities.
Question 2: write all five other trigonometric ratios of ∠A in terms of sec A.
Question 3: choose the correct option (MCQ) for four trigonometric simplifications and justify each choice.
Question 4: prove ten trigonometric identities (parts i-x) covering a wide range of techniques.

How to Solve Exercise 8.3 Question by Question

The whole exercise turns on one core idea: use the three Pythagorean identities as a toolkit. Every question in Exercise 8.3 can be solved by picking the right identity and expressing everything in terms of sin and cos when stuck. The approach below works on every sub-part.

Question	What it asks	Key technique
Q1	Express sin A, sec A, tan A in terms of cot A	Reciprocal square identities csc²A = 1 + cot²A
Q2	All five ratios in terms of sec A	Chain: cos A = 1/sec A then Pythagoras
Q3 (i)	9sec²A - 9tan²A	Factor 9 and use sec²A - tan²A = 1
Q3 (ii-iv)	Three MCQ simplifications	Convert to sin/cos; difference of squares
Q4 (i-x)	Ten identity proofs	Work from complex side; rationalize roots; combine fractions

Quick Tip: In identity proofs, always work on one side only. Never cross-multiply or move terms across the equality sign while proving. Start from the left-hand side (or whichever is more complex) and reduce it to the other.

Three Pythagorean Identities Used in Exercise 8.3

Every question in this exercise connects back to these three identities. Students who know them well can solve Exercise 8.3 quickly and without confusion.

Identity	Form used in Exercise 8.3	Questions where it appears
sin²A + cos²A = 1	Rewrite sin²A = 1 - cos²A or vice versa	Q2, Q3(ii), Q3(iii), Q4(i-ix)
sec²A - tan²A = 1	Factor expressions, simplify products	Q1, Q2, Q3(i), Q3(iv), Q4(v)
csc²A - cot²A = 1	Express sin in terms of cot; prove csc-cot identities	Q1, Q4(v), Q4(viii)

All three come from the basic identity sin²A + cos²A = 1. Divide through by cos²A to get the secant-tangent identity, or divide by sin²A to get the cosecant-cotangent identity.
Reciprocal relations: tan A = sin A / cos A, cot A = cos A / sin A, sec A = 1 / cos A, csc A = 1 / sin A.

Proof Techniques Students Need for Exercise 8.3

Question 4 has ten sub-parts covering every major technique. Knowing which method to pick saves time in the exam. Most lost marks here come from not knowing which tool to reach for first.

Combine fractions over a common denominator: Parts (ii) and (iii). Merge two fractions, simplify the numerator using sin² + cos² = 1, and the answer appears cleanly.
Difference of squares / factor method: Parts (i), (iv), (vi), (vii). Replace 1 - cos² with (1 - cos)(1 + cos) and cancel the matching factor.
Rationalise under the root: Part (vi). Multiply top and bottom by (1 + sin A) to clear the root and get a clean fraction.
Expand and group: Part (viii). Expand squared brackets, then group sin² + cos² = 1 inside; the 7 + tan² + cot² form falls out.
Divide the proof LHS by sin A: Part (v). Introduces csc and cot quickly, then use the identity to cancel the awkward denominator.

Watch Out: In Q4 part (x), the middle expression ((1 - tan A) / (1 - cot A))² simplifies to (-tan A)² = tan²A. The minus sign inside gives a positive square, which is a common trap that costs marks.

Exercise 8.3 Marks and CBSE Board Trends for Class 10

Introduction to Trigonometry typically carries 7 to 8 marks in the CBSE Class 10 paper. Exercise 8.3 contributes the bulk of those marks through identity proofs and ratio simplifications.

Question type	Where it appears in the board paper	Typical marks
Ratio in terms of another (Q1, Q2 style)	Short-answer slots	2 to 3
MCQ simplification (Q3 style)	1-mark objective section	1
Identity proof short (one-step)	3-mark or 4-mark questions	3 to 4
Identity proof multi-step (Q4 complex parts)	4-mark or 5-mark questions	4 to 5

These solutions follow the 2026-27 NCERT exactly, so the working you practise here matches what the board paper rewards.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry: All Exercises

Chapter 8 has three exercises. The table below links each one to its own step-by-step solution page.

Exercise	Topics covered	Solutions page
Exercise 8.1	Trigonometric ratios from right triangles; ratios from given sides	Exercise 8.1 Solutions
Exercise 8.2	Specific angle values (0, 30, 45, 60, 90 degrees) and complementary angles	Exercise 8.2 Solutions
Exercise 8.3 (this page)	Trigonometric identities, ratio rewriting, ten identity proofs	Exercise 8.3 Solutions
Full chapter	All exercises combined	Chapter 8 NCERT Solutions (all exercises)

Other Resources for Chapter 8 Introduction to Trigonometry Class 10 Maths

Use the table below to move between other resources for Chapter 8, other chapters, and other RTs. Each link opens the matching Collegedunia page.

Resource	Open page
Chapter 8 full NCERT Solutions	Introduction to Trigonometry Class 10 NCERT Solutions
Exercise 8.1	Exercise 8.1 Solutions
Exercise 8.2	Exercise 8.2 Solutions
Revision notes	Introduction to Trigonometry Class 10 Notes
Formula sheet	Introduction to Trigonometry Class 10 Formula Sheet
NCERT book PDF	Introduction to Trigonometry Class 10 NCERT Book PDF
Handwritten notes	Introduction to Trigonometry Class 10 Handwritten Notes
Exemplar solutions	Introduction to Trigonometry Class 10 Exemplar Solutions
Chapter 7 NCERT Solutions	Coordinate Geometry NCERT Solutions
Chapter 9 NCERT Solutions	Some Applications of Trigonometry NCERT Solutions

All NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3 with Step-by-Step Solutions

Exercise 8.3

Q 8.1

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Q 8.2

Write all the other trigonometric ratios of ∠ A in terms of sec A.

Q 8.3

Choose the correct option. Justify your choice:
(i) 9sec² A-9tan² A=   (A) 1 (B) 9 (C) 8 (D) 0
(ii) (1+tanθ+secθ)(1+cotθ-cscθ)=   (A) 0 (B) 1 (C) 2 (D) -1
(iii) (sec A+tan A)(1-sin A)=   (A) sec A (B) sin A (C) csc A (D) cos A
(iv) 1+tan² A1+cot² A=   (A) sec² A (B) -1 (C) cot² A (D) tan² A.

Q 8.4

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
(i) (cscθ-cotθ)²=1-cosθ1+cosθ
(ii) cos A1+sin A+1+sin Acos A=2sec A
(iii) tanθ1-cotθ+cotθ1-tanθ=1+secθ
(iv) 1+sec Asec A=sin² A1-cos A
(v) cos A-sin A+1cos A+sin A-1=csc A+cot A, using csc² A=1+cot² A
(vi) √1+sin A1-sin A=sec A+tan A
(vii) sinθ-2sin³θ2cos³θ-cosθ=tanθ
(viii) (sin A+csc A)²+(cos A+sec A)²=7+tan² A+cot² A
(ix) (csc A-sin A)(sec A-cos A)=1tan A+cot A
(x) (1+tan² A1+cot² A)=(1-tan A1-cot A)²=tan² A.

Student Feedback

Out of 21,500 students surveyed before the 2026 boards, 88% said Exercise 8.3 identity proofs became easier once they always started from the more complex side and used sin/cos substitution as a fallback. Marks on proof questions improved a lot after practising all ten sub-parts of Question 4.

Source: Collegedunia Class 10 student survey, 2026 board batch.

Introduction to Trigonometry Class 10 Maths Exercise 8.3 NCERT Solutions FAQs

Ques. How many questions are there in Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3?

Ans. Exercise 8.3 has 4 questions. Question 1 asks for three ratios in terms of cot A, Question 2 asks for all ratios in terms of sec A, Question 3 has four MCQ simplifications, and Question 4 requires students to prove ten trigonometric identities.

Ques. What identities are needed for Exercise 8.3?

Ans. The three Pythagorean identities cover almost the whole exercise: sin²A + cos²A = 1, sec²A - tan²A = 1, and csc²A - cot²A = 1. You also need the reciprocal definitions: sec = 1/cos, csc = 1/sin, tan = sin/cos, cot = cos/sin.

Ques. How do I prove identities in Exercise 8.3 Question 4?

Ans. Always work on one side only, starting from the more complex side. Never move terms to the other side or cross-multiply. The main techniques are: combine fractions over a common denominator, factor as a difference of squares, rationalise a square root by multiplying by the conjugate, or convert everything to sin and cos when stuck.

Ques. Why does part (x) of Question 4 give a negative sign in the middle expression?

Ans. When you simplify (1 - tan A)/(1 - cot A), rewrite cot A as 1/tan A. The denominator becomes (tan A - 1)/tan A, so the whole fraction becomes -tan A. The minus sign disappears when you square it, giving tan²A, which equals the other two expressions in Question 4(x).

Ques. Are these Exercise 8.3 solutions based on the 2026-27 syllabus?

Ans. Yes. These solutions follow the current 2026-27 CBSE syllabus for Class 10 Mathematics. All four questions in Exercise 8.3 are fully retained in the latest 2026-27 NCERT edition of Chapter 8 Introduction to Trigonometry.