Chapter 8: Introduction to Trigonometry

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

1. Q1Short answer3 marksAsked: 2023, 2024

   Prove that : sin theta / (1 + cos theta) + (1 + cos theta) / sin theta = 2 cosec theta

   

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   Model answer · Proving trigonometric identities

   1. 1.To prove: sin θ/(1 + cos θ) + (1 + cos θ)/sin θ = 2 cosec θ. Taking the LHS and writing both fractions over the common denominator sin θ (1 + cos θ): LHS = [sin²θ + (1 + cos θ)²] / [sin θ (1 + cos θ)] = (sin²θ + 1 + cos²θ + 2 cos θ) / (sin θ (1 + cos θ))1 mark
   2. 2.Using the trigonometric identity sin² θ + cos² θ = 1, the numerator becomes 1 + 1 + 2 cos θ: = (2 + 2 cos θ) / (sin θ (1 + cos θ))1 mark
   3. 3.Taking 2 common in the numerator, 2(1 + cos θ), and cancelling (1 + cos θ) from the numerator and the denominator: = 2 / sin θ0.5 marks
   4. 4.Since cosec θ = 1/sin θ: = 2 cosec θ = RHS Hence proved.0.5 marks

   • Write out the expansion (1 + cos θ)² = 1 + cos²θ + 2 cos θ in the numerator before you simplify it.
   • State the identity sin² θ + cos² θ = 1 at the step where the numerator becomes 2 + 2 cos θ.
   • Finish with "= 2 cosec θ = RHS", using cosec θ = 1/sin θ.

   Marks split as in the board’s published marking scheme

2. Q2Short answer3 marksAsked: 2023, 2024

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

   

   Medha Ma’am

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   Model answer · Proving trigonometric identities

   1. 1.Given: sin θ + cos θ = p and sec θ + cosec θ = q. L.H.S. = q(p² − 1) = (sec θ + cosec θ)[(sin θ + cos θ)² − 1] Using sec θ = 1/cos θ and cosec θ = 1/sin θ, and expanding the square: = [1/cos θ + 1/sin θ][sin² θ + cos² θ + 2 sin θ cos θ − 1]1 mark
   2. 2.Taking the LCM in the first bracket, and using the identity sin² θ + cos² θ = 1 in the second bracket: = [(sin θ + cos θ)/(sin θ cos θ)][1 + 2 sin θ cos θ − 1]1 mark
   3. 3.= [(sin θ + cos θ)/(sin θ cos θ)] × 2 sin θ cos θ Cancelling sin θ cos θ: = 2(sin θ + cos θ)0.5 marks
   4. 4.Since sin θ + cos θ = p, = 2p = R.H.S. Hence proved: q(p² − 1) = 2p.0.5 marks

   • Write the substitution step "(sec θ + cosec θ)[(sin θ + cos θ)² − 1]" before you expand anything.
   • When you replace sin² θ + cos² θ with 1, say that you are using the identity sin² θ + cos² θ = 1.
   • End by putting sin θ + cos θ = p to get "= 2p = R.H.S."

   Marks split as in the board’s published marking scheme

3. Q3Case-based4 marksAsked: 2025

   Rahim and Nadeem are two friends whose plots are adjacent to each other. Rahim's son made a drawing of the plots with necessary details. It is decided that Rahim will fence the triangular plot ABC and Nadeem will fence along the sides AF, FE and BE. Observe the diagram carefully and answer the following questions : (Use sqrt(2) = 1.41 and sqrt(3) = 1.73) (i) Find length BC. (ii) Find length AG. (iii) (a) Calculate perimeter of triangle ABC. OR (iii) (b) Calculate length of (AF + FE + EB).

   

   Medha Ma’am

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   Model answer · Trigonometric ratios, case study

   1. 1.(i) In the right triangle with the 45° angle, the side adjacent to angle 45° is 50 m and BC is the hypotenuse. cos 45° = side adjacent to angle 45° / hypotenuse = 50/BC 1/√2 = 50/BC BC = 50√2 = 50 × 1.41 BC = 70.5 m1 mark
   2. 2.(ii) In the right triangle with the 45° angle at G's side, the side opposite to angle 45° is 38 m and AG is the side adjacent to angle 45°. tan 45° = side opposite to angle 45° / side adjacent to angle 45° = 38/AG 1 = 38/AG AG = 38 m1 mark
   3. 3.(iii) (a) In the right triangle with the 60° angle, the side opposite to angle 60° is 78 m and AC is the hypotenuse. sin 60° = side opposite to angle 60° / hypotenuse = 78/AC √3/2 = 78/AC AC = 156/√3 = 52√3 = 52 × 1.73 AC = 89.96 m1 mark
   4. 4.The perimeter of triangle ABC is the sum of the lengths of its boundary: BC = 70.5 m, AC = 89.96 m, and the remaining side made up of the parts 38 m, 50 m and 40 m. Perimeter of Δ ABC = 70.5 + 89.96 + 38 + 50 + 40 Perimeter of Δ ABC = 288.46 m1 mark

   • Write the ratio with its sides first, for example "cos 45° = 50/BC", and then put in the value from the table.
   • Put in √2 = 1.41 and √3 = 1.73 only at the last step, as in "AC = 52√3 = 52 × 1.73".
   • Give every length in metres: "BC = 70.5 m", "AG = 38 m", "288.46 m".

   Marks split as in the board’s published marking scheme

4. Q4Short answer3 marksAsked: 2026

   (A) If x = h + a cos theta, y = k + b sin theta, then prove that : ((x - h)/a)^2 + ((y - k)/b)^2 = 1 OR (B) Prove that : tan A / (1 + sec A) - tan A / (1 - sec A) = 2 cosec A

   How the model answer begins

   1. Solving (A). Given: x = h + a cos θ and y = k + b sin θ. From the first equation, subtract h from both sides: x − h = a cos θ Divide both sides by a: (x − h)/a = cos θ ... (1)

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5. Q5Short answer3 marksAsked: 2026

   (A) If x = h + a cos theta, y = k + b sin theta, then prove that : ((x - h)/a)^2 + ((y - k)/b)^2 = 1 OR (B) Prove that : (tan A / (1 + sec A)) - (tan A / (1 - sec A)) = 2 cosec A

6. Q6Short answer3 marksAsked: 2026

   (a) Prove that: (sec^3 theta)/(sec^2 theta - 1) + (cosec^3 theta)/(cosec^2 theta - 1) = sec theta . cosec theta (sec theta + cosec theta) OR (b) If (sec alpha)/(cosec beta) = p and (tan alpha)/(cosec beta) = q, then prove that (p^2 - q^2) sec^2 alpha = p^2.

7. Q7Short answer3 marksAsked: 2026

   (a) If sin theta + cos theta = sqrt(3), then prove that tan theta + cot theta = 1. OR (b) Prove that: (sin A + sec A)^2 + (cos A + cosec A)^2 = (1 + sec A cosec A)^2

8. Q8Short answer3 marksAsked: 2026

   (a) Prove that (cot A - cos A) / (cot A + cos A) = (sec A - tan A) / (sec A + tan A) OR (b) Prove that (cosec A - sin A) (sec A - cos A) = 1 / (tan A + cot A)

   How the model answer begins

   1. (a) We have to prove that (cot A − cos A)/(cot A + cos A) = (sec A − tan A)/(sec A + tan A). LHS = (cot A − cos A)/(cot A + cos A) Put cot A = cos A/sin A and multiply the numerator and the denominator by sin A: LHS =…

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9. Q9Short answer3 marksAsked: 2026

   (a) Prove that: ((sec theta + tan theta)^2 - 1) / ((sec theta + tan theta)^2 + 1) = sin theta OR (b) Prove that: (tan A / (1 - cot A)) + (cot A / (1 - tan A)) = 1 + sec A cosec A.

   How the model answer begins

   1. (a) To prove: ((sec θ + tan θ)² − 1) / ((sec θ + tan θ)² + 1) = sin θ Expanding the squares in the numerator and the denominator: LHS = (sec²θ + tan²θ + 2 sec θ tan θ − 1) / (sec²θ + tan²θ + 2 sec θ tan θ + 1)

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10. Q10Short answer3 marksAsked: 2025

    Prove the following trigonometric identity : cos theta / (1 + sin theta) + (1 + sin theta) / cos theta = 2 sec theta

    How the model answer begins

    1. Solution: LHS = cos θ/(1 + sin θ) + (1 + sin θ)/cos θ Taking the LCM (1 + sin θ) cos θ of the two denominators: LHS = [cos²θ + (1 + sin θ)²] / [(1 + sin θ) cos θ]

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11. Q11Short answer3 marksAsked: 2025

    Prove the following trigonometric identity : (1 - sin theta / cos theta) + (cos theta / 1 - sin theta) = 2 sec theta

    How the model answer begins

    1. To prove: (1 − sin θ)/cos θ + cos θ/(1 − sin θ) = 2 sec θ LHS = (1 − sin θ)/cos θ + cos θ/(1 − sin θ) We add the two fractions over their LCM, cos θ (1 − sin θ): LHS = [(1 − sin θ)² + cos² θ] / [cos θ (1 − sin θ)]

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12. Q12Short answer3 marksAsked: 2025

    26. Prove the following trigonometric identity : (cos A - 2 cos^3 A) / (2 sin^3 A - sin A) = cot A

    How the model answer begins

    1. To prove: (cos A − 2 cos³ A) / (2 sin³ A − sin A) = cot A. We start from the LHS. Taking cos A common from the numerator and sin A common from the denominator: LHS = (cos A − 2 cos³ A) / (2 sin³ A − sin A) = cos A (1 −…

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13. Q13Short answer3 marksAsked: 2025

    Prove that : (1 + cosec A) / cosec A = cos^2 A / (1 - sin A)

    How the model answer begins

    1. Since cosec A = 1/sin A, write the left-hand side in terms of sin A: LHS = (1 + cosec A) / cosec A = (1 + 1/sin A) / (1/sin A)

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14. Q14Short answer3 marksAsked: 2025

    Prove the following trigonometric identity : (tan theta / (1 + cot theta)) + (cot theta / (1 + tan theta)) = tan theta + cot theta - 1

    How the model answer begins

    1. To prove the trigonometric identity, start with the left-hand side. LHS = tan θ / (1 + cot θ) + cot θ / (1 + tan θ)

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15. Q15Short answer3 marksAsked: 2025

    Prove that (cosec A + sin A)^2 + (sec A + cos A)^2 = 7 + tan^2 A + cot^2 A.

    How the model answer begins

    1. We expand the LHS using (a + b)^2 = a^2 + b^2 + 2ab: LHS = (cosec A + sin A)^2 + (sec A + cos A)^2 = cosec^2 A + sin^2 A + 2 cosec A sin A + sec^2 A + cos^2 A + 2 cos A sec A

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16. Q16Short answer3 marksAsked: 2025

    Prove that : (cos theta / (1 - tan theta)) + (sin theta / (1 - cot theta)) = cos theta + sin theta.

    How the model answer begins

    1. Using tan θ = sin θ / cos θ and cot θ = cos θ / sin θ: LHS = cos θ / (1 − sin θ / cos θ) + sin θ / (1 − cos θ / sin θ)

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17. Q17Short answer3 marksAsked: 2025

    Prove that : (sin θ + sec θ)^2 + (cos θ + cosec θ)^2 = (1 + sec θ cosec θ)^2.

    How the model answer begins

    1. LHS = (sin θ + sec θ)² + (cos θ + cosec θ)² Expanding each square using (a + b)² = a² + b² + 2ab: LHS = sin²θ + sec²θ + 2 sin θ sec θ + cos²θ + cosec²θ + 2 cos θ cosec θ

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18. Q18Short answer3 marksAsked: 2025

    Prove that (tan theta / (1 - cot theta)) + (cot theta / (1 - tan theta)) = sec theta cosec theta + 1.

    How the model answer begins

    1. To prove this, write the LHS in terms of sin θ and cos θ, using tan θ = sin θ / cos θ and cot θ = cos θ / sin θ: LHS = (sin θ / cos θ) / (1 − cos θ / sin θ) + (cos θ / sin θ) / (1 − sin θ / cos θ)

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19. Q19Short answer3 marksAsked: 2025

    Prove that (1 + 1/tan^2 θ)(1 + 1/cot^2 θ) = 1/(sin^2 θ - sin^4 θ)

    How the model answer begins

    1. To prove: (1 + 1/tan²θ)(1 + 1/cot²θ) = 1/(sin²θ − sin⁴θ). In each bracket, take the LCM: LHS = [(1 + tan²θ)/tan²θ] × [(1 + cot²θ)/cot²θ]

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20. Q20Short answer3 marksAsked: 2025

    Prove that (sin theta) / (1 + cos theta) + (1 + cos theta) / (sin theta) = 2 cosec theta.

    How the model answer begins

    1. LHS = sin θ / (1 + cos θ) + (1 + cos θ) / sin θ Taking the LCM of the denominators, (1 + cos θ) sin θ: LHS = [sin²θ + (1 + cos θ)²] / [(1 + cos θ) sin θ]

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118 more questions of this chapter, each with its model answer, in the Ask Medha app.

Objective questions with the board’s key (10)

1. 1. CBSE 2026

   If sin theta = a/b, then cos theta is equal to :

   • (A) b / sqrt(b^2 - a^2)
   • (B) b / a
   • (C) sqrt(b^2 - a^2) / b ✓
   • (D) a / sqrt(b^2 - a^2)

   Answer: (C)

2. 2. CBSE 2026

   If cos A = 1/2, then the value of sin^2 A + 2 cos^2 A is :

   • (A) 3/2
   • (B) 5/4 ✓
   • (C) -1
   • (D) 1/2

   Answer: (B)

3. 3. CBSE 2026

   If cos y = 0, then what is the value of 1/2 cos y/2 ?

   • (A) 0
   • (B) 1/2
   • (C) 1/sqrt(2)
   • (D) 1/(2sqrt(2)) ✓

   Answer: (D)

4. 4. CBSE 2026

   If cos A = 4/5, then the value of tan A is :

   • (A) 3/5
   • (B) 3/4 ✓
   • (C) 4/3
   • (D) 5/3

   Answer: (B)

5. 5. CBSE 2026

   If 2 sin A = 1, then the value of tan A + cot A is :

   • (A) sqrt(3)
   • (B) 4/sqrt(3) ✓
   • (C) sqrt(3)/2
   • (D) 1

   Answer: (B)

6. 6. CBSE 2026

   If cot theta = 3, then the value of cos theta is :

   • (A) 1/3
   • (B) 1/sqrt(10)
   • (C) 3/sqrt(10) ✓
   • (D) sqrt(10)/3

   Answer: (C)

7. 7. CBSE 2026

   The value of (1/2 tan^2 45 degree - cos^2 60 degree) is :

   • (A) 0
   • (B) - 1/2
   • (C) 1/4 ✓
   • (D) - 1/4

   Answer: (C)

8. 8. CBSE 2026

   Assertion (A) : tan 2 theta is not defined at theta = 45 degree. Reason (R) : sin 90 degree not equal to cos 90 degree.

   • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
   • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). ✓
   • (C) Assertion (A) is true, but Reason (R) is false.
   • (D) Assertion (A) is false, but Reason (R) is true.

   Answer: (B)

9. 9. CBSE 2026

   Given that sin 2alpha = sqrt(3)/2, the value of sin 3alpha is :

   • (A) (3*sqrt(3))/4
   • (B) 1/2
   • (C) 1 ✓
   • (D) sqrt(3)/4

   Answer: (C)

10. 10. CBSE 2026

    The value of (1/3 cot^2 30° - 1/2 sec^2 60°) is :

    • (A) - 1 ✓
    • (B) - 2
    • (C) 5/8
    • (D) 7/8

    Answer: (A)

Questions parents and students ask

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

In the CBSE papers of 2023–2026, this chapter carried about 7 marks and about 4 questions a year; by marks it ranks 3 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 138 written questions of this chapter, with answers, are in the Ask Medha app.

Do the answers follow the board’s marking scheme?

Yes. CBSE publishes its marking schemes; every model answer here is written point by point to the marks the board’s own scheme gives, and checked by a second, independent model.