How many marks does "Introduction to Trigonometry" carry in Bihar Board Class 10 Maths?

In the Bihar Board papers of 2021–2026, this chapter carried about 17 marks and about 25 questions a year; by marks it ranks 1 in Maths.

Which are the most important (VVI) questions of "Introduction to Trigonometry"?

The 20 questions above are from the board’s real papers, each with the years it was asked; the most-asked come first with their full model answers. All 57 written questions of this chapter, with answers, are in the AskMedha app.

Do the answers follow the board’s marking scheme?

BSEB does not publish a marking scheme. Each answer is written point by point on a structure derived from past papers and checked by a second, independent model; it is never called the board’s official scheme.

Bihar Board Class 10 · Maths · Matric exam 2027

Chapter 8: Introduction to Trigonometry

This chapter’s questions from Bihar Board (BSEB)’s real papers of 2017–2026 — the 20 asked most often, 3 with the full model answer, 17 with how the answer begins — and objective questions with the board’s key. All 57 questions of this chapter, with answers, are in the app.

marks a year (about): 17
questions a year (about): 25
rank in Maths: #1

From 6 years of papers, 2021–2026 · full chapter-wise marks

Chapter 8 notes (PDF) →

VVI questions — the ones the board asked most (20)

Q1Long answer5 marksAsked: 2020, 2022
Prove that (1 + cot theta - cosec theta)(1 + tan theta + sec theta) = 2.
Medha Ma’am
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Model answer · trigonometric identities, proof
1. 1.Writing cot θ = cos θ/sin θ, cosec θ = 1/sin θ, tan θ = sin θ/cos θ, sec θ = 1/cos θ: LHS = (1 + cos θ/sin θ − 1/sin θ)(1 + sin θ/cos θ + 1/cos θ).1 mark
2. 2.Taking the LCM: = [(sin θ + cos θ − 1)/sin θ] × [(sin θ + cos θ + 1)/cos θ].1 mark
3. 3.Using (a − b)(a + b) = a² − b²: = [(sin θ + cos θ)² − 1]/(sin θ · cos θ).1 mark
4. 4.(sin θ + cos θ)² − 1 = sin²θ + cos²θ + 2 sin θ cos θ − 1 = 1 + 2 sin θ cos θ − 1 = 2 sin θ cos θ.1 mark
5. 5.So LHS = 2 sin θ cos θ/(sin θ cos θ) = 2 = RHS. Hence proved.1 mark
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
Q2Long answer5 marksAsked: 2020, 2022
Prove that (cosec A - sin A) (sec A - cos A) = 1 / (tan A + cot A)
Medha Ma’am
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Model answer · trigonometric identity proof
1. 1.To prove (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A), we take the Left Hand Side (LHS) and write cosec A and sec A in terms of sin A and cos A. Since cosec A = 1/sin A and sec A = 1/cos A, the LHS becomes (1/sin A - sin A)(1/cos A - cos A). Taking the terms inside each bracket to a common denominator, this becomes ((1 - sin^2 A)/sin A)((1 - cos^2 A)/cos A). Using the identity cos^2 A + sin^2 A = 1, we have 1 - sin^2 A = cos^2 A and 1 - cos^2 A = sin^2 A.1 mark
2. 2.Substituting 1 - sin^2 A = cos^2 A and 1 - cos^2 A = sin^2 A in the above expression, the LHS simplifies to (cos^2 A/sin A) × (sin^2 A/cos A). This is obtained by multiplying the two fractions, keeping cos^2 A and sin^2 A in the numerators as found above, with sin A and cos A remaining in the respective denominators.1 mark
3. 3.Solving (cos^2 A/sin A) × (sin^2 A/cos A) further, one cos A from the numerator cancels with one cos A in the denominator, and one sin A from the numerator cancels with one sin A in the denominator. This leaves sin A × cos A. Therefore, LHS = sin A cos A.1 mark
4. 4.Now we take the Right Hand Side (RHS), which is 1/(tan A + cot A). Writing tan A = sin A/cos A and cot A = cos A/sin A, the RHS becomes 1/(sin A/cos A + cos A/sin A).1 mark
5. 5.Taking the terms in the denominator of the RHS to a common denominator, sin A/cos A + cos A/sin A becomes (sin^2 A + cos^2 A)/(sin A cos A). So the RHS becomes 1/((sin^2 A + cos^2 A)/(sin A cos A)). Using the identity sin^2 A + cos^2 A = 1, this reduces to 1/(1/(sin A cos A)), which equals sin A cos A. Thus RHS = sin A cos A. Since LHS = sin A cos A and RHS = sin A cos A, we get LHS = RHS. Hence, (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A) is proved.1 mark
- Write cosec A = 1/sin A and sec A = 1/cos A at the start of the LHS working, and clearly show cos^2 A + sin^2 A = 1 being used to get 1 - sin^2 A = cos^2 A.
- Show the multiplication step (cos^2 A/sin A) × (sin^2 A/cos A) explicitly before cancelling to reach sin A cos A.
- On the RHS, write tan A = sin A/cos A and cot A = cos A/sin A, then use sin^2 A + cos^2 A = 1 to simplify the denominator to sin A cos A, ending with LHS = RHS.
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
Q3Short answer2 marksAsked: 2020, 2022
If 3 tan theta = 4, then find the value of (3 sin theta + 2 cos theta) / (3 sin theta - 2 cos theta).
Medha Ma’am
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Model answer · trigonometric ratios, evaluation
1. 1.Given 3 tan θ = 4, so tan θ = 4/3.0.5 marks
2. 2.Dividing the numerator and denominator of (3 sin θ + 2 cos θ)/(3 sin θ − 2 cos θ) by cos θ, we get (3 sin θ/cos θ + 2)/(3 sin θ/cos θ − 2) = (3 tan θ + 2)/(3 tan θ − 2).0.5 marks
3. 3.Substituting tan θ = 4/3 in (3 tan θ + 2)/(3 tan θ − 2): (3 × 4/3 + 2)/(3 × 4/3 − 2) = (4 + 2)/(4 − 2)0.5 marks
4. 4.On simplification, (4 + 2)/(4 − 2) = 6/2. Therefore, the value of (3 sin θ + 2 cos θ)/(3 sin θ − 2 cos θ) = 3.0.5 marks
- Write tan θ = 4/3 clearly as the first step before anything else.
- Show the division step by cos θ to get the expression in terms of tan θ, writing (3 tan θ + 2)/(3 tan θ − 2).
- After substitution, show the simplification clearly ending in the final numeric answer 3.
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
Q4Long answer5 marksAsked: 2025
Prove that (sec theta - tan theta) / (sec theta + tan theta) = 1 + 2 tan^2 theta - 2 sec theta . tan theta.
How the model answer begins
1. To prove that (sec θ – tan θ)/(sec θ + tan θ) = 1 + 2 tan² θ – 2 sec θ . tan θ, we start with the left hand side LHS = (sec θ – tan θ)/(sec θ + tan θ). We multiply and divide this expression by (sec θ – tan θ), so that…
The full answer from Medha Ma’am →
Q5Long answer5 marksAsked: 2026
Prove that ((1 - tan theta)/(1 - cot theta))^2 = (1 + tan^2 theta)/(1 + cot^2 theta) = tan^2 theta.
How the model answer begins
1. Using sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ: (1 + tan²θ)/(1 + cot²θ) = sec²θ/cosec²θ.
The full answer from Medha Ma’am →
Q6Long answer5 marksAsked: 2026
Prove that ((1 - tan theta)/(1 - cot theta))^2 = (1 + tan^2 theta)/(1 + cot^2 theta) = tan^2 theta.
How the model answer begins
1. 1 − cot θ = 1 − 1/tan θ = (tan θ − 1)/tan θ.
The full answer from Medha Ma’am →
Q7Long answer5 marksAsked: 2022
Prove that (sin theta - cos theta + 1) / (sin theta + cos theta - 1) = 1 / (sec theta - tan theta).
How the model answer begins
1. To prove that (sin theta - cos theta + 1) / (sin theta + cos theta - 1) = 1 / (sec theta - tan theta), we first convert the LHS in terms of sec theta and tan theta by dividing the numerator and denominator by cos theta.…
The full answer from Medha Ma’am →
Q8Long answer5 marksAsked: 2023
Prove that tan 7°. tan 23°. tan 60°. tan 67°. tan 83° = sqrt(3).
How the model answer begins
1. We are given tan 7°. tan 23°. tan 60°. tan 67°. tan 83°. We know that tan (90° – A) = cot A. So, tan 83° = tan (90° – 7°) = cot 7°. Thus we can pair tan 7° with tan 83° and write tan 7°. tan 83° as tan 7°. cot 7°.
The full answer from Medha Ma’am →
Q9Long answer5 marksAsked: 2023
Prove that tan 7°. tan 23°. tan 60°. tan 67°. tan 83° = sqrt(3).
How the model answer begins
1. Using the identity tan (90° – θ) = cot θ, we write tan 83° as tan 83° = tan (90° – 7°) = cot 7°. So the term tan 7° remains as tan 7°, and tan 83° is written as cot 7°.
The full answer from Medha Ma’am →
Q10Long answer5 marksAsked: 2021
Prove that tan theta / (1 - cot theta) + cot theta / (1 - tan theta) = 1 + tan theta + cot theta.
How the model answer begins
1. To prove that tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + tan θ + cot θ, we start with the LHS and write cot θ in terms of tan θ using cot θ = 1/tan θ. So tan θ/(1 − cot θ) becomes tan θ/(1 − 1/tan θ). The second term…
The full answer from Medha Ma’am →
Q11Long answer5 marksAsked: 2021
Prove that (1 + sec θ)/sec θ = sin^2 θ/(1 - cos θ)
How the model answer begins
1. To prove that (1 + sec θ)/sec θ = sin^2 θ/(1 - cos θ), we start with the Left Hand Side (LHS). Since sec A = 1/cos A, we write sec θ = 1/cos θ. So the LHS (1 + sec θ)/sec θ becomes (1 + 1/cos θ) / (1/cos θ).
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Q12Short answer2 marksAsked: 2025
If tan theta = 5/12 then find the value of sin theta + cos theta.
How the model answer begins
1. Since tan theta = 5/12, consider a right-angled triangle in which the side opposite to theta (perpendicular) = 5 and the side adjacent to theta (base) = 12. By Pythagoras theorem, hypotenuse = root(perpendicular^2 +…
The full answer from Medha Ma’am →
Q13Short answer2 marksAsked: 2025
If sin 3A = cos(A - 26°), where 3A is an acute angle, then find the value of A.
How the model answer begins
1. We are given that sin 3A = cos (A – 26°), where 3A is an acute angle. Since sin A = cos (90° – A) for all values of angle A lying between 0° and 90°, we can write sin 3A = cos (90° – 3A). Substituting this in the given…
The full answer from Medha Ma’am →
Q14Short answer2 marksAsked: 2025
Prove that tan 7° . tan 60° . tan 83° = sqrt(3).
How the model answer begins
1. tan 83° = tan(90° − 7°) = cot 7°.
The full answer from Medha Ma’am →
Q15Short answer2 marksAsked: 2025
Prove that sqrt((1 + cos theta) / (1 - cos theta)) = (1 + cos theta) / sin theta.
How the model answer begins
1. To prove sqrt((1+cos theta)/(1-cos theta)) = (1+cos theta)/sin theta, we start with LHS = sqrt((1+cos theta)/(1-cos theta)). Rationalizing the denominator by multiplying numerator and denominator inside the square root…
The full answer from Medha Ma’am →
Q16Short answer2 marksAsked: 2025
Prove that tan 9° . tan 27° = cot 63° . cot 81°.
How the model answer begins
1. We know that tan (90° – A) = cot A. Writing tan 9° as tan (90° – 81°), we get tan 9° = cot 81°.
The full answer from Medha Ma’am →
Q17Short answer2 marksAsked: 2025
If cos A = 4/5 then find the values of cot A and cosec A.
How the model answer begins
1. Given cos A = 4/5, consider a right triangle ABC, right-angled at B, with ∠A the angle whose cosine is taken. Since cos A = AB/AC = 4/5, let AB = 4k and AC = 5k, where k is a positive number. By the Pythagoras theorem,…
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Q18Short answer2 marksAsked: 2026
Evaluate : (sec 39 degree / cosec 51 degree) + (sin 67 degree / cos 23 degree) - cot 45 degree - cos 0 degree.
How the model answer begins
1. sec 39° can be written using complementary angles as sec 39° = sec (90° – 51°) = cosec 51°. So sec 39°/cosec 51° = cosec 51°/cosec 51° = 1.
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Q19Short answer2 marksAsked: 2026
If 4tan theta = 3 then find the value of sin theta cos theta.
How the model answer begins
1. Given 4 tan theta = 3, we get tan theta = 3/4.
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Q20Short answer2 marksAsked: 2026
Express sin theta and cos theta in the form of tan theta.
How the model answer begins
1. The value of sin θ in terms of tan θ is obtained using the identity sin² A + cos² A = 1. Dividing both sides by cos² A gives sec² A = 1 + tan² A, so cos θ = 1/√(1 + tan² θ). Since tan θ = sin θ/cos θ, we get sin θ = tan…
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37 more questions of this chapter, each with its model answer, in the AskMedha app.

Objective questions with the board’s key (10)

1. Bihar Board 2026
If sin theta = 45/53 then cos theta =
- (A) 53/28
- (B) 28/53 ✓
- (C) 28/45
- (D) 45/28
Answer: (B)
2. Bihar Board 2026
sin 29° / cos 61° =
- (A) 90
- (B) 0
- (C) 1 ✓
- (D) 32
Answer: (C)
3. Bihar Board 2026
Which of the following options is not equal to unity ?
- (A) cosec^2 theta - cot^2 theta
- (B) sec^2 theta - tan^2 theta
- (C) cos^2 theta - sin^2 theta ✓
- (D) cos^2 theta + sin^2 theta
Answer: (C)
4. Bihar Board 2026
The value of cos theta in the form of sin theta is
- (A) sqrt(1 - sin^2 theta) ✓
- (B) sqrt(1 + sin^2 theta)
- (C) sqrt(sin^2 theta - 1)
- (D) none of these
Answer: (A)
5. Bihar Board 2026
If sec theta = cosec theta then theta =
- (A) 90°
- (B) 45° ✓
- (C) 30°
- (D) 75°
Answer: (B)
6. Bihar Board 2026
sin^2 30° + cos^2 60° - cot^2 45° =
- (A) 1/2
- (B) -1/2 ✓
- (C) 1
- (D) 0
Answer: (B)
7. Bihar Board 2026
If sqrt(3) tan theta = 1 then theta =
- (A) 30° ✓
- (B) 45°
- (C) 60°
- (D) 75°
Answer: (A)
8. Bihar Board 2026
If cot A + 1/cot A = 2 then tan^2 A + 1/tan^2 A =
- (A) 0
- (B) 1
- (C) 2 ✓
- (D) 4
Answer: (C)
9. Bihar Board 2026
sin^4 theta - cos^4 theta =
- (A) cos^2theta - sin^2theta
- (B) 2cos^2theta - 1
- (C) 1 + 2cos^2theta
- (D) 2sin^2theta - 1 ✓
Answer: (D)
10. Bihar Board 2026
If sin theta + sin^2 theta = 1 then cos^2 theta + cos^4 theta =
- (A) 0
- (B) 1 ✓
- (C) 2
- (D) 4
Answer: (B)

Questions parents and students ask

How many marks does "Introduction to Trigonometry" carry in Bihar Board Class 10 Maths?: In the Bihar Board papers of 2021–2026, this chapter carried about 17 marks and about 25 questions a year; by marks it ranks 1 in Maths.
Which are the most important (VVI) questions of "Introduction to Trigonometry"?: The 20 questions above are from the board’s real papers, each with the years it was asked; the most-asked come first with their full model answers. All 57 written questions of this chapter, with answers, are in the AskMedha app.
Do the answers follow the board’s marking scheme?: BSEB does not publish a marking scheme. Each answer is written point by point on a structure derived from past papers and checked by a second, independent model; it is never called the board’s official scheme.

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