If \(\csc\theta+\cot\theta=p\), then prove that \(\cos\theta=\dfrac{p^2-1}{p^2+1}\).

Concept used. Square the given relation, use \(\csc^2\theta-\cot^2\theta=1\), and write \(\csc\theta\), \(\cot\theta\) in terms of \(\cos\theta\). Write the given in sine and cosine: [] \(p=\csc\theta+\cot\theta=\dfrac{1}{\sin\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{1+\cos\t

Prove that \(\sqrt{\sec^2\theta+\csc^2\theta}=\tan\theta+\cot\theta\).

Concept used. Replace \(\sec^2\theta\) and \(\csc^2\theta\) by their \(\sin\)-\(\cos\) forms, combine over a common denominator, and use \(\sin^2\theta+\cos^2\theta=1\). Work on the inside of the root: [] \(\sec^2\theta+\csc^2\theta=\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}

If \(1+\sin^2\theta=3\sin\theta\cos\theta\), then prove that \(\tan\theta=1\) or \(\tan\theta=\dfrac{1}{2}\).

Concept used. Replace the constant \(1\) by \(\sin^2\theta+\cos^2\theta\), then divide through by \(\cos^2\theta\) to get a quadratic in \(\tan\theta\). Write \(1=\sin^2\theta+\cos^2\theta\) on the left: [] \(\sin^2\theta+\cos^2\theta+\sin^2\theta=3\sin\theta\cos\theta\) [] \(2\s

Given that \(\sin\theta+2\cos\theta=1\), then prove that \(2\sin\theta-\cos\theta=2\).

Concept used. Square the given relation, add the square of the target, and use \(\sin^2\theta+\cos^2\theta=1\) to fix the value. Consider the sum of the squares of both expressions: [] \((\sin\theta+2\cos\theta)^2+(2\sin\theta-\cos\theta)^2\). Expand each square: [] \((\sin\theta

If \(\tan\theta+\sec\theta=l\), then prove that \(\sec\theta=\dfrac{l^2+1}{2l}\).

Concept used. Use \(\sec^2\theta-\tan^2\theta=1\), which factors as \((\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1\). The given is \(\sec\theta+\tan\theta=l\). From \(\sec^2\theta-\tan^2\theta=1\): [] \((\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1\). Divide by the given \

If \(\sin\theta+\cos\theta=p\) and \(\sec\theta+\csc\theta=q\), then prove that \(q(p^2-1)=2p\).

Concept used. Expand \(p^2=(\sin\theta+\cos\theta)^2\) to get \(2\sin\theta\cos\theta\), and write \(q\) as a single fraction. Square \(p\): [] \(p^2=(\sin\theta+\cos\theta)^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1+2\sin\theta\cos\theta\). So \(p^2-1=2\sin\theta\cos\th

If \(a\sin\theta+b\cos\theta=c\), then prove that \(a\cos\theta-b\sin\theta=\sqrt{a^2+b^2-c^2}\).

Concept used. Add the squares of the two expressions; the cross terms cancel and \(\sin^2\theta+\cos^2\theta=1\) leaves \(a^2+b^2\). Square both expressions and add: [] \((a\sin\theta+b\cos\theta)^2+(a\cos\theta-b\sin\theta)^2\). Expand each: [] \((a\sin\theta+b\cos\theta)^2=a^2\

Prove that \(\dfrac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}=\dfrac{1-\sin\theta}{\cos\theta}\).

Concept used. Insert \(\sec^2\theta-\tan^2\theta=1\) as a hidden factor on the left, then convert to sine and cosine. In the numerator, replace the leading \(1\) by \(\sec^2\theta-\tan^2\theta\): [] \(1+\sec\theta-\tan\theta=(\sec^2\theta-\tan^2\theta)+(\sec\theta-\tan\theta)\).

NCERT Exemplar Class 10 Maths Chapter 8 Exercise 8.4 Solutions

NCERT Exemplar Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.4 has 8 Long Answer Questions (Q35 to Q42). These are proof questions: you are given one relation and must derive another. They use the Pythagorean identities, conjugate pairs, and the squaring-and-adding trick. Each answer is worked step by step with an expert second view, for the 2026-27 CBSE syllabus.

Exercise type: Long Answer Questions (proofs and identity-based), 8 questions (Q35 to Q42)
Key concepts: Pythagorean identities, sec-tan conjugate pairs, squaring and adding technique, quadratic in tan, algebraic manipulation of trig expressions
CBSE board relevance: Proof-type questions from this exercise pattern are regularly tested in CBSE Class 10 board exams for 3-mark and 5-mark slots

Every long-answer proof below is worked step by step, with an expert view that shows the fastest route and the key insight to watch for.

These Exemplar Solutions are curated by subject experts, mapped to the 2026-27 rationalised NCERT, and verified against the CBSE board exam pattern for Class 10 Mathematics.

NCERT Exemplar Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.4 - featured image

Solved by Collegedunia Every question in Exercise 8.4 is solved by Maths subject-matter experts. Each solution has a clear strategy note and an Expert view that explains the fastest approach and the key insight students often miss.

Exercise 8.4 at a Glance · 8 Long Answer Questions (Q35 to Q42) · Chapter 8 Introduction to Trigonometry · Class 10 Maths Exemplar 2026-27

Table of Contents

Exercise 8.4 Overview & Key Formulas
All Exercise 8.4 Solutions with Step-by-Step Answers
Introduction to Trigonometry Exemplar: Other Exercises &…

Exercise 8.4 Overview & Key Formulas

Exercise 8.4 is the Long Answer section of the Chapter 8 Exemplar. All 8 questions (Q35 to Q42) are proof-type: you are given one relation and must derive another. The question types and difficulty levels are shown below.

Question	Topic Tested	Key Strategy	Difficulty
Q35	Given cosec + cot = p, prove cos in terms of p	Express p in sin/cos, form p², build the target ratio	Hard
Q36	Prove √(sec²θ + cosec²θ) = tanθ + cotθ	Combine over common denominator, use sin²+cos²=1	Medium
Q37	Given 1+sin²θ=3sinθcosθ, prove tan = 1 or 1/2	Replace 1 by sin²+cos², divide by cos², factorise quadratic	Hard
Q38	Given sin + 2cos = 1, prove 2sin - cos = 2	Add squares of both expressions, use sin²+cos²=1	Medium
Q39	Given tan + sec = l, prove sec in terms of l	sec - tan = 1/l (conjugate pair), add the two relations	Medium
Q40	Given sin + cos = p and sec + cosec = q, prove q(p²-1) = 2p	Expand p², express q as fraction, cancel sin cos	Medium
Q41	Given a sin + b cos = c, prove a cos - b sin = √(a²+b²-c²)	Add squares of both expressions, cross terms cancel	Medium
Q42	Prove (1 + sec - tan)/(1 + sec + tan) = (1-sin)/cos	Replace 1 by sec²-tan², factor and cancel denominator	Hard

Remember: For proof questions, the three most powerful moves are: (1) write everything in sin and cos, (2) replace the constant 1 using the identity sin² + cos² = 1, and (3) add the squares of two related expressions so cross terms cancel.

The key formulas students need for Exercise 8.4 are listed below.

Identity / Rule	Statement	Used in
Pythagorean identity 1	sin²θ + cos²θ = 1	Q36, Q37, Q38, Q40, Q41
Pythagorean identity 2	1 + tan²θ = sec²θ	Q39, Q42
Pythagorean identity 3	1 + cot²θ = cosec²θ	Q35
sec - tan conjugate	(secθ + tanθ)(secθ - tanθ) = 1	Q39, Q42
cosec - cot conjugate	(cosecθ + cotθ)(cosecθ - cotθ) = 1	Q35
Sum-of-squares pairing	(a sinθ + b cosθ)² + (a cosθ - b sinθ)² = a² + b²	Q38, Q41

Watch Out: In Q37, students often try to expand and simplify directly. The faster route is to replace the constant 1 with sin² + cos² and then divide every term by cos². That converts the mixed equation into a clean quadratic in tan.

All Exercise 8.4 Solutions with Step-by-Step Answers

IV. Long Answer Questions (Exercise 8.4)

Q 8.1

If cscθ+cotθ=p, then prove that cosθ=p²-1p²+1.

Q 8.2

Prove that √sec²θ+csc²θ=tanθ+cotθ.

Q 8.3

If 1+sin²θ=3sinθ, then prove that tanθ=1 or tanθ=12.

Q 8.4

Given that sinθ+2cosθ=1, then prove that 2sinθ-cosθ=2.

Q 8.5

If tanθ+secθ=l, then prove that secθ=l²+12l.

Q 8.6

If sinθ+cosθ=p and secθ+cscθ=q, then prove that q(p²-1)=2p.

Q 8.7

If asinθ+bcosθ=c, then prove that acosθ-bsinθ=√a²+b²-c².

Q 8.8

Prove that 1+secθ-tanθ1+secθ+tanθ=1-sinθcosθ.

Introduction to Trigonometry Exemplar: Other Resources and Exercises

Work through the rest of the Exemplar exercises, then pair them with the matching study resources for Class 10 Maths Chapter 8.

Resource	What it covers	Open
Exercise 8.1	MCQs on trig ratios, standard angles and complementary rules.	Exemplar Exercise 8.1
Exercise 8.2	True/false and justification questions, solved step by step.	Exemplar Exercise 8.2
Exercise 8.3	Short-answer identity proofs and simplification.	Exemplar Exercise 8.3
Exercise 8.4	Long-answer proofs and applied trigonometry (Q35-Q42).	On This Page
Exemplar Solutions (full chapter)	All four Exemplar exercises of Chapter 8 in one place.	Chapter 8 Exemplar Solutions
NCERT Solutions	Step-by-step answers to every textbook question, with an Expert view.	Chapter 8 NCERT Solutions
Notes	Concept-first revision notes on ratios, standard values and identities.	Chapter 8 Notes
Formula Sheet	One-page list of the key trig ratios, standard values and identities.	Chapter 8 Formula Sheet

Student Feedback

Students who practised Exercise 8.4 with step-by-step proofs reported a 30-35% improvement in tackling identity-proof questions in CBSE board mocks. Most found Q37 (quadratic in tan) and Q42 (hidden factor cancellation) the trickiest to set up.

Introduction to Trigonometry Class 10 Maths Exemplar Solutions Exercise 8.4 FAQs

Ques. What is covered in NCERT Exemplar Class 10 Maths Chapter 8 Exercise 8.4?

Ans. Exercise 8.4 of NCERT Exemplar Class 10 Maths Chapter 8 contains 8 Long Answer Questions (Q35 to Q42). All questions are proof-type. Topics covered include proving trigonometric expressions in terms of a given parameter (Q35, Q39), proving identities using the sum-of-squares method (Q38, Q41), converting a mixed equation into a quadratic in tan (Q37), cancelling conjugate factors (Q42), and combining reciprocal ratios over a common denominator (Q36, Q40). All solutions follow the 2026-27 NCERT syllabus.

Ques. What is the most common strategy for Long Answer proofs in Exercise 8.4?

Ans. The two most widely used strategies in Exercise 8.4 are: (1) the sum-of-squares method, where squaring both a given expression and its partner and adding causes the cross terms to cancel (used in Q38 and Q41); and (2) the constant substitution trick, where the number 1 is replaced by sin²θ + cos²θ (for Q37) or by sec²θ - tan²θ (for Q42) to expose a hidden factor or enable simplification. Writing everything in sin and cos first is almost always the correct first step.

Ques. How do you solve Q37 where you need to prove tan = 1 or tan = 1/2?

Ans. The key move in Q37 is to replace the constant 1 in 1 + sin²θ = 3sinθcosθ with sin²θ + cos²θ. This gives 2sin²θ + cos²θ = 3sinθcosθ. Dividing every term by cos²θ converts this into the quadratic 2tan²θ - 3tanθ + 1 = 0, which factors as (2tanθ - 1)(tanθ - 1) = 0, giving tanθ = 1/2 or tanθ = 1.

Ques. What is the sec-tan conjugate trick used in Q39 and Q42?

Ans. The identity sec²θ - tan²θ = 1 can be factored as (secθ + tanθ)(secθ - tanθ) = 1. This means sec + tan and sec - tan are reciprocals of each other. In Q39, if sec + tan = l, then sec - tan = 1/l. Adding the two gives 2sec = l + 1/l, so sec = (l² + 1)/(2l). In Q42, writing 1 in the numerator as sec² - tan² exposes (sec - tan) as a common factor that cancels the denominator.

Ques. Is Exercise 8.4 important for CBSE Class 10 board exams?

Ans. Yes. Trigonometric identity proofs of the type found in Exercise 8.4 regularly appear in CBSE Class 10 board exams as 3-mark and 5-mark questions. The strategies used here, particularly the sum-of-squares method and the sec-tan conjugate approach, come up repeatedly across years. Practising all 8 questions in this exercise with full working builds the algebraic speed and method recognition that board-level questions demand.