Chapter 8 · Class 10 CBSE

Introduction to Trigonometry 2-Mark Questions Class 10: CBSE PYQs with Answers

Q: Can an angle equation have two valid answers?

Yes. For example, sin 2A = √3/2 permits 2A = 60° or 120° when the stated interval permits both. Check the given angle restrictions before rejecting a value. The two possibilities give A = 30° or 60°.

Trigonometry two-mark questions ask you to evaluate standard angles, find missing ratios or angles, and prove identities. Practise 84 question groups below, with both alternatives of each OR pair solved in short, separate steps.

By Siddharth Agrawal, Lead Maths Tutor, TutorVertex · Updated 5 November 2026
Solutions prepared with Team TutorVertex

121 selected board-paper appearances · 10 official-sample appearances · 46 OR pairs · 130 worked answers

Trigonometry MCQs Assertion–Reason PYQs Case-study PYQs

Key results

Useful trigonometric relationships
Use	Relationship
Change a squared ratio	sin²A + cos²A = 1 sec²A − tan²A = 1 cosec²A − cot²A = 1
Convert ratios	tan A = sin A / cos A; cot A = cos A / sin A sec A = 1 / cos A; cosec A = 1 / sin A
Find a missing side	Hypotenuse² = sum of the squares of the two legs
Choose an angle	Use the interval printed in the question; check each permitted value

Topic

Level

Source

Showing all 84 question groups

Questions on this page · Show list

Standard values and numerical verification

Q1 · Basic
Q2 · Basic
Q3 · Basic · OR pair
Q4 · Standard · OR pair
Q5 · Standard · OR pair
Q6 · Basic
Q7 · Basic
Q8 · Basic
Q9 · Basic · OR pair
Q10 · Basic
Q11 · Basic
Q12 · Standard · OR pair
Q13 · Basic · OR pair
Q14 · Standard
Q15 · Standard · OR pair
Q16 · Basic · OR pair
Q17 · Basic
Q18 · Standard · OR pair
Q19 · Standard · OR pair
Q20 · Basic · OR pair
Q21 · Standard · OR pair
Q22 · Standard · OR pair
Q23 · Basic
Q24 · Basic
Q25 · Basic
Q26 · Standard
Q27 · Standard
Q28 · Standard
Q29 · Standard
Q30 · Standard · OR pair
Q31 · Basic
Q32 · Standard · OR pair
Q33 · Basic
Q34 · Standard
Q35 · Basic · OR pair

Ratios and right triangles

Q36 · Basic
Q37 · Basic · OR pair
Q38 · Standard · OR pair
Q39 · Standard · OR pair
Q40 · Standard · OR pair
Q41 · Basic · OR pair
Q42 · Standard
Q43 · Standard · OR pair
Q44 · Basic
Q45 · Basic · OR pair
Q46 · Basic
Q47 · Basic · OR pair
Q48 · Basic · OR pair
Q49 · Standard
Q50 · Basic
Q51 · Standard · OR pair
Q52 · Basic
Q53 · Basic
Q54 · Standard

Finding angles

Q55 · Basic
Q56 · Basic
Q57 · Basic
Q58 · Standard · OR pair
Q59 · Standard · OR pair
Q60 · Basic · OR pair
Q61 · Basic · OR pair
Q62 · Basic · OR pair
Q63 · Standard · OR pair
Q64 · Standard · OR pair
Q65 · Standard · OR pair
Q66 · Basic
Q67 · Standard

Identities and algebraic relations

Q68 · Standard
Q69 · Standard · OR pair
Q70 · Standard · OR pair
Q71 · Standard · OR pair
Q72 · Standard · OR pair
Q73 · Basic · OR pair
Q74 · Standard · OR pair
Q75 · Basic · OR pair
Q76 · Basic
Q77 · Standard · OR pair
Q78 · Standard · OR pair
Q79 · Standard
Q80 · Basic
Q81 · Basic
Q82 · Standard · OR pair
Q83 · Standard · OR pair
Q84 · Standard · OR pair

Standard values and numerical verification

Basic2 marks

Q1.Evaluate: 5 cosec²45° − 3 sin²90° + 5 cos 0°.

Show answer and solution

cosec 45° = √2, sin 90° = 1, cos 0° = 1.

Given expression = 5 × 2 − 3 × 1 + 5 × 1.

Answer: 12

Appeared in: 2023 · Basic · Main

Basic2 marks

Q2.Find sin A cos B + cos A sin B, where A = 30° and B = 45°.

Show answer and solution

Substitute the standard values.

= ( $\frac{1}{2}$ )( $\frac{1}{\sqrt{2}}$ ) + ( $\frac{\sqrt{3}}{2}$ )( $\frac{1}{\sqrt{2}}$ ).

Answer: (√6 + √2)/4

Appeared in: 2024 · Basic · Main

Basic2 marks · internal OR

Q3.(a) Evaluate: 2 tan²45° + cos²30° − sin²90°.

Show answer and solution

= 2(1) + $\frac{3}{4}$ − 1.

Answer: $\frac{7}{4}$

(b) Verify cos 2A = (1 − tan²A)/(1 + tan²A), when A = 30°.

Show answer and solution

LHS = cos 60° = $\frac{1}{2}$ .

RHS = (1 − $\frac{1}{3}$ )/(1 + $\frac{1}{3}$ ) = ( $\frac{2}{3}$ )/( $\frac{4}{3}$ ) = $\frac{1}{2}$ .

Answer: LHS = RHS; verified

Appeared in: 2025 · Basic · Main

Standard2 marks · internal OR

Q4.(a) Evaluate: (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + cos²30°).

Show answer and solution

Denominator = $\frac{1}{4}$ + $\frac{3}{4}$ = 1.

Numerator = $\frac{5}{4}$ + $\frac{16}{3}$ − 1 = $\frac{67}{12}$ .

Answer: $\frac{67}{12}$

(b) If A and B are acute angles, sin(A − B) = 0 and 2 cos(A + B) − 1 = 0, find A and B.

Show answer and solution

A − B = 0 and cos(A + B) = $\frac{1}{2}$ .

A + B = 60°.

Adding gives 2A = 60°.

Answer: A = 30°, B = 30°

Appeared in: 2023 · Standard · Main

Standard2 marks · internal OR

Q5.(a) Evaluate 2 sec²θ + 3 cosec²θ − 2 sin θ cos θ, when θ = 45°.

Show answer and solution

= 2(2) + 3(2) − 2( $\frac{1}{\sqrt{2}}$ )( $\frac{1}{\sqrt{2}}$ ).

Answer: 9

(b) If sin θ − cos θ = 0, find sin⁴θ + cos⁴θ.

Show answer and solution

sin θ = cos θ, so sin²θ = cos²θ = $\frac{1}{2}$ .

sin⁴θ + cos⁴θ = $\frac{1}{4}$ + $\frac{1}{4}$ .

Answer: $\frac{1}{2}$

Appeared in: 2023 · Standard · Main

Basic2 marks

Q6.Evaluate: tan²60° − 2 cosec²30° − 2 tan²30°.

Show answer and solution

= 3 − 2(4) − 2( $\frac{1}{3}$ ).

Answer: − $\frac{17}{3}$

Appeared in: 2023 · Basic · Main

Basic2 marks

Q7.Evaluate: 2(sin²45° + cot²30°) − 6(cos²45° − tan²30°).

Show answer and solution

= 2( $\frac{1}{2}$ + 3) − 6( $\frac{1}{2}$ − $\frac{1}{3}$ ).

= 7 − 1.

Answer: 6

Appeared in: 2023 · Basic · Main

Basic2 marks

Q8.Evaluate: ( $\frac{3}{2}$ )tan²30° − 2 cos²90° − ( $\frac{1}{2}$ )cosec²30°.

Show answer and solution

= ( $\frac{3}{2}$ )( $\frac{1}{3}$ ) − 0 − ( $\frac{1}{2}$ )(4).

Answer: − $\frac{3}{2}$

Appeared in: 2023 · Basic · Main

Basic2 marks · internal OR

Q9.(a) Verify tan(B − A) = (tan B − tan A)/(1 + tan A tan B), for A = 30° and B = 60°.

Show answer and solution

LHS = tan 30° = $\frac{1}{\sqrt{3}}$ .

RHS = (√3 − $\frac{1}{\sqrt{3}}$ )/(1 + 1) = $\frac{1}{\sqrt{3}}$ .

Answer: LHS = RHS; verified

(b) Evaluate: sin²60° − 2 cos²45° + ( $\frac{1}{2}$ )cosec²30°.

Show answer and solution

= $\frac{3}{4}$ − 2( $\frac{1}{2}$ ) + ( $\frac{1}{2}$ )(4).

Answer: $\frac{7}{4}$

Appeared in: 2023 · Basic · Supplementary

Basic2 marks

Q10.Evaluate: 3 sec²30° cosec 30° + tan²60° tan²45°.

Show answer and solution

= 3( $\frac{4}{3}$ )(2) + 3(1).

Answer: 11

Appeared in: 2024 · Basic · Main

Basic2 marks

Q11.Evaluate: 5 sin²45° − sec 60° cot²30°.

Show answer and solution

= 5( $\frac{1}{2}$ ) − 2(3).

Answer: − $\frac{7}{2}$

Appeared in: 2024 · Basic · Main

Standard2 marks · internal OR

Q12.(a) If x cos 60° + y cos 0° + sin 30° − cot 45° = 5, find x + 2y.

Show answer and solution

x/2 + y + $\frac{1}{2}$ − 1 = 5.

x/2 + y = $\frac{11}{2}$ .

Multiply by 2.

Answer: x + 2y = 11

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

Show answer and solution

= 3/( $\frac{3}{4}$ + $\frac{3}{4}$ ) = 3/( $\frac{3}{2}$ ).

Answer: 2

Appeared in: 2025 · Standard · Main

Basic2 marks · internal OR

Q13.(a) Evaluate: (5 sin²45° − 3 tan²30°)/(2 sec²30°).

Show answer and solution

= ( $\frac{5}{2}$ − 1)/( $\frac{8}{3}$ ) = ( $\frac{3}{2}$ )( $\frac{3}{8}$ ).

Answer: $\frac{9}{16}$

(b) Verify tan(A − B) = (tan A − tan B)/(1 + tan A tan B), for A = 60°, B = 30°.

Show answer and solution

LHS = tan 30° = $\frac{1}{\sqrt{3}}$ .

RHS = (√3 − $\frac{1}{\sqrt{3}}$ )/(1 + 1) = $\frac{1}{\sqrt{3}}$ .

Answer: LHS = RHS; verified

Appeared in: 2026 · Basic · Main

Standard2 marks

Q14.If tan A = 1 and tan B = √3, find cos A cos B + sin A sin B.

Show answer and solution

For acute angles, A = 45°, B = 60°.

= ( $\frac{1}{\sqrt{2}}$ )( $\frac{1}{2}$ ) + ( $\frac{1}{\sqrt{2}}$ )( $\frac{\sqrt{3}}{2}$ ).

Answer: (√6 + √2)/4

Appeared in: 2023 · Standard · Supplementary

Standard2 marks · internal OR

Q15.(a) If 4 cot²45° − sec²60° + sin²60° + p = $\frac{3}{4}$ , find p.

Show answer and solution

4(1) − 4 + $\frac{3}{4}$ + p = $\frac{3}{4}$ .

Answer: p = 0

(b) If cos A + cos²A = 1, find sin²A + sin⁴A.

Show answer and solution

sin²A = 1 − cos²A = cos A.

sin²A + sin⁴A = cos A + cos²A.

Answer: 1

Appeared in: 2023 · Standard · Main

Basic2 marks · internal OR

Q16.(a) Evaluate: (sin 30° + tan 45°)/(sec 30° + cot 45°).

Show answer and solution

= ( $\frac{3}{2}$ )/( $\frac{2}{\sqrt{3}}$ + 1) = 3√3/[2(2 + √3)].

Rationalise using 2 − √3.

Answer: (6√3 − 9)/2

(b) Verify sin(A + B) = sin A cos B + cos A sin B, for A = 30°, B = 60°.

Show answer and solution

LHS = sin 90° = 1.

RHS = ( $\frac{1}{2}$ )( $\frac{1}{2}$ ) + ( $\frac{\sqrt{3}}{2}$ )( $\frac{\sqrt{3}}{2}$ ) = 1.

Answer: LHS = RHS; verified

Appeared in: 2023 · Basic · Main

Basic2 marks

Q17.Evaluate: ( $\frac{3}{4}$ )tan²30° − 3 sin²60° + 1/sin²45°.

Show answer and solution

= ( $\frac{3}{4}$ )( $\frac{1}{3}$ ) − 3( $\frac{3}{4}$ ) + 2.

= $\frac{1}{4}$ − $\frac{9}{4}$ + 2.

Answer: 0

Appeared in: 2023 · Basic · Supplementary

Standard2 marks · internal OR

Q18.(a) Evaluate: 2 sin²30° sec 60° + tan²60°.

Show answer and solution

= 2( $\frac{1}{4}$ )(2) + 3.

Answer: 4

(b) If 2 sin(A + B) = √3 and cos(A − B) = 1, where A, B and A + B lie between 0° and 90°, find A and B.

Show answer and solution

A + B = 60° and A − B = 0°.

Add the equations: 2A = 60°.

Answer: A = 30°, B = 30°

Appeared in: 2024 · Standard · Main

Standard2 marks · internal OR

Q19.(a) Evaluate: (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + sin²60°).

Show answer and solution

Denominator = $\frac{1}{4}$ + $\frac{3}{4}$ = 1.

Numerator = $\frac{5}{4}$ + $\frac{16}{3}$ − 1 = $\frac{67}{12}$ .

Answer: $\frac{67}{12}$

(b) If sin(A − B) = $\frac{1}{2}$ and cos(A + B) = $\frac{1}{2}$ , where 0° < A + B ≤ 90° and A > B, find A and B.

Show answer and solution

A − B = 30°, A + B = 60°.

Adding gives 2A = 90°.

Answer: A = 45°, B = 15°

Appeared in: 2024 · Standard · Main

Basic2 marks · internal OR

Q20.(a) Evaluate: cos 45°/(tan 30° + sin 60°).

Show answer and solution

= ( $\frac{1}{\sqrt{2}}$ )/( $\frac{1}{\sqrt{3}}$ + $\frac{\sqrt{3}}{2}$ ).

= ( $\frac{1}{\sqrt{2}}$ )(2√ $\frac{3}{5}$ ).

Answer: $\frac{\sqrt{6}}{5}$

(b) Verify sin 2A = 2 tan A/(1 + tan²A), when A = 30°.

Show answer and solution

LHS = sin 60° = $\frac{\sqrt{3}}{2}$ .

RHS = ( $\frac{2}{\sqrt{3}}$ )/(1 + $\frac{1}{3}$ ) = $\frac{\sqrt{3}}{2}$ .

Answer: LHS = RHS; verified

Appeared in: 2025 · Basic · Main

Standard2 marks · internal OR

Q21.(a) Evaluate: 2√2 cos 45° sin 30° + 2√3 cos 30°.

Show answer and solution

= 2√2( $\frac{1}{\sqrt{2}}$ )( $\frac{1}{2}$ ) + 2√3( $\frac{\sqrt{3}}{2}$ ).

= 1 + 3.

Answer: 4

(b) Verify sin(A + B) = sin A cos B + cos A sin B, for A = 60°, B = 30°.

Show answer and solution

LHS = sin 90° = 1.

RHS = ( $\frac{\sqrt{3}}{2}$ )( $\frac{\sqrt{3}}{2}$ ) + ( $\frac{1}{2}$ )( $\frac{1}{2}$ ) = 1.

Answer: LHS = RHS; verified

Appeared in: 2024 · Standard · Main

Standard2 marks · internal OR

Q22.(a) Evaluate: (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + sin²60°).

Show answer and solution

Denominator = 1.

Numerator = $\frac{5}{4}$ + $\frac{16}{3}$ − 1.

Answer: $\frac{67}{12}$

(b) Prove: cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.

Show answer and solution

LHS = [cos²A + (1 + sin A)²]/[cos A(1 + sin A)].

= [2(1 + sin A)]/[cos A(1 + sin A)].

= 2/cos A.

Answer: LHS = 2 sec A = RHS

Appeared in: 2023 · Standard · Main

Basic2 marks

Q23.Evaluate: sin 30° cos 60° + cos 30° sin 60° − cot 45°.

Show answer and solution

= ( $\frac{1}{2}$ )( $\frac{1}{2}$ ) + ( $\frac{\sqrt{3}}{2}$ )( $\frac{\sqrt{3}}{2}$ ) − 1.

Answer: 0

Appeared in: 2024 · Basic · Main

Basic2 marks

Q24.Evaluate: sin²30° + cos²45° − cos 0° tan 45°.

Show answer and solution

= $\frac{1}{4}$ + $\frac{1}{2}$ − 1.

Answer: − $\frac{1}{4}$

Appeared in: 2024 · Basic · Main

Basic2 marks

Q25.Find x, if x = sin 60° cos 30° + cos 60° sin 30°.

Show answer and solution

x = ( $\frac{\sqrt{3}}{2}$ )( $\frac{\sqrt{3}}{2}$ ) + ( $\frac{1}{2}$ )( $\frac{1}{2}$ ).

Answer: x = 1

Appeared in: 2024 · Basic · Main

Standard2 marks

Q26.Evaluate: (cos 45° + sin 60°)/(sec 30° + cosec 30°).

Show answer and solution

= ( $\frac{1}{\sqrt{2}}$ + $\frac{\sqrt{3}}{2}$ )/( $\frac{2}{\sqrt{3}}$ + 2).

Multiply numerator and denominator by √3.

Answer: (√6 + 3)/[4(1 + √3)]

Appeared in: 2024 · Standard · Main

Standard2 marks

Q27.Evaluate: 5 tan 60°/[(sin²60° + cos²60°)tan 30°].

Show answer and solution

sin²60° + cos²60° = 1.

Expression = 5√3/( $\frac{1}{\sqrt{3}}$ ).

Answer: 15

Appeared in: 2024 · Standard · Main

Standard2 marks

Q28.Evaluate: (2 tan 30° sec 60° tan 45°)/(1 − sin²60°).

Show answer and solution

= [2( $\frac{1}{\sqrt{3}}$ )(2)(1)]/( $\frac{1}{4}$ ).

= $\frac{16}{\sqrt{3}}$ .

Answer: 16√ $\frac{3}{3}$

Appeared in: 2024 · Standard · Main

Standard2 marks

Q29.Evaluate: 2 sin²30° tan 60° − 3 cos²60° sec 30°.

Show answer and solution

= 2( $\frac{1}{4}$ )√3 − 3( $\frac{1}{4}$ )( $\frac{2}{\sqrt{3}}$ ).

= $\frac{\sqrt{3}}{2}$ − $\frac{\sqrt{3}}{2}$ .

Answer: 0

Appeared in: 2026 · Standard · Second board examination

Standard2 marks · internal OR

Q30.(a) If 2 cosec²30° + x sin²60° − ( $\frac{3}{4}$ )tan²30° = 10, find x.

Show answer and solution

8 + 3x/4 − $\frac{1}{4}$ = 10.

3x/4 = $\frac{9}{4}$ .

Answer: x = 3

(b) If cos A = $\frac{2}{5}$ , find 4 + 4 tan²A.

Show answer and solution

1 + tan²A = sec²A.

4 + 4 tan²A = 4 sec²A = 4( $\frac{25}{4}$ ).

Answer: 25

Appeared in: 2026 · Standard · Second board examination

Basic2 marks

Q31.Verify sin²α cot β = cos β tan α, for α = 45°, β = 30°.

Show answer and solution

LHS = ( $\frac{1}{2}$ )√3 = $\frac{\sqrt{3}}{2}$ .

RHS = ( $\frac{\sqrt{3}}{2}$ )(1) = $\frac{\sqrt{3}}{2}$ .

Answer: LHS = RHS; verified

Appeared in: 2026 · Basic · Second board examination

Standard2 marks · internal OR

Q32.(a) Evaluate: (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + cos²30°).

Show answer and solution

Denominator = 1.

Numerator = $\frac{5}{4}$ + $\frac{16}{3}$ − 1.

Answer: $\frac{67}{12}$

(b) Prove: 1 + cot²α/(1 + cosec α) = cosec α.

Show answer and solution

LHS = 1 + (cosec²α − 1)/(1 + cosec α).

= 1 + cosec α − 1.

Answer: LHS = cosec α = RHS

Appeared in: 2026 · Standard · Main

Basic2 marks

Q33.Evaluate: sin²60° + 2 tan 45° − cos²30°.

Show answer and solution

= $\frac{3}{4}$ + 2 − $\frac{3}{4}$ .

Answer: 2

Appeared in: 2023–24 · Basic · Official sample

Standard2 marks

Q34.Evaluate: (2 sin²60° − tan²30°)/sec²45°.

Show answer and solution

= ( $\frac{3}{2}$ − $\frac{1}{3}$ )/2 = ( $\frac{7}{6}$ )/2.

Answer: $\frac{7}{12}$

Appeared in: 2024–25 · Standard · Official sample

Basic2 marks · internal OR

Q35.(a) Evaluate: 2 sin 30° tan 60° − 3 cos²60° sec²30°.

Show answer and solution

= 2( $\frac{1}{2}$ )√3 − 3( $\frac{1}{4}$ )( $\frac{4}{3}$ ).

Answer: √3 − 1

(b) If sin x = $\frac{7}{25}$ , where x is acute, find sin x cos x(tan x + cot x).

Show answer and solution

tan x + cot x = (sin²x + cos²x)/(sin x cos x).

Multiplying by sin x cos x gives 1.

Answer: 1

Appeared in: 2025–26 · Basic · Official sample

Ratios and right triangles

Basic2 marks

Q36.In triangle ABC, right-angled at A, tan C = √3. Find sin B + cos C − cos²B.

Show answer and solution

C = 60° and B = 30°.

Expression = $\frac{1}{2}$ + $\frac{1}{2}$ − $\frac{3}{4}$ .

Answer: $\frac{1}{4}$

Appeared in: 2024 · Basic · Main

Basic2 marks · internal OR

Q37.(a) If sec A = $\frac{25}{7}$ , find cosec A and tan A.

Show answer and solution

Take hypotenuse = 25k and adjacent side = 7k.

Opposite side = √(625k² − 49k²) = 24k.

cosec A = 25k/(24k), tan A = 24k/(7k).

Answer: cosec A = $\frac{25}{24}$ ; tan A = $\frac{24}{7}$

(b) Verify sin(A + B) = cos A sin B + cos B sin A, for A = 60°, B = 30°.

Show answer and solution

LHS = sin 90° = 1.

RHS = ( $\frac{1}{2}$ )( $\frac{1}{2}$ ) + ( $\frac{\sqrt{3}}{2}$ )( $\frac{\sqrt{3}}{2}$ ) = 1.

Answer: LHS = RHS; verified

Appeared in: 2025 · Basic · Main

Standard2 marks · internal OR

Q38.(a) Using sin(A − B) = sin A cos B − cos A sin B, find sin 15°.

Show answer and solution

Put A = 45°, B = 30°.

sin 15° = ( $\frac{1}{\sqrt{2}}$ )( $\frac{\sqrt{3}}{2}$ ) − ( $\frac{1}{\sqrt{2}}$ )( $\frac{1}{2}$ ).

Answer: (√6 − √2)/4

(b) If sin A = y, express cos A and tan A in terms of y.

Show answer and solution

In the acute-angle setting, cos A is positive.

cos²A = 1 − y².

tan A = sin A/cos A.

Answer: cos A = √(1 − y²); tan A = y/√(1 − y²)

Appeared in: 2025 · Standard · Main

Standard2 marks · internal OR

Q39.(a) If tan A = √2 − 1, find tan A/(1 + tan²A).

Show answer and solution

1 + tan²A = 1 + (√2 − 1)² = 4 − 2√2.

Expression = (√2 − 1)/(4 − 2√2).

Multiply numerator and denominator by 2 + √2.

Answer: $\frac{\sqrt{2}}{4}$

(b) If cosec θ = $\frac{13}{12}$ , find cot θ + tan θ.

Show answer and solution

Take hypotenuse = 13k, opposite = 12k.

Adjacent = √(169k² − 144k²) = 5k.

cot θ + tan θ = $\frac{5}{12}$ + $\frac{12}{5}$ .

Answer: $\frac{169}{60}$

Appeared in: 2026 · Standard · Second board examination

Standard2 marks · internal OR

Q40.(a) If tan A = $\frac{24}{7}$ , find sin A + cos A.

Show answer and solution

Take opposite = 24k, adjacent = 7k.

Hypotenuse = √(576k² + 49k²) = 25k.

sin A + cos A = $\frac{24}{25}$ + $\frac{7}{25}$ .

Answer: $\frac{31}{25}$

(b) If cot A = $\frac{7}{8}$ , find [(1 + sin A)(1 − sin A)]/[(1 + cos A)(1 − cos A)].

Show answer and solution

Expression = (1 − sin²A)/(1 − cos²A).

= cos²A/sin²A = cot²A.

Answer: $\frac{49}{64}$

Appeared in: 2026 · Standard · Main

Basic2 marks · internal OR

Q41.(a) If sec A = √2 and tan B = √3, find 2 sin A cos B.

Show answer and solution

For acute angles, A = 45°, B = 60°.

2 sin A cos B = 2( $\frac{1}{\sqrt{2}}$ )( $\frac{1}{2}$ ).

Answer: $\frac{1}{\sqrt{2}}$

(b) Evaluate: (4 cos³60° + cosec 30°)/tan²30°.

Show answer and solution

= [4( $\frac{1}{8}$ ) + 2]/( $\frac{1}{3}$ ).

= ( $\frac{5}{2}$ )(3).

Answer: $\frac{15}{2}$

Appeared in: 2026 · Basic · Main

Standard2 marks

Q42.If tan A = √3, where A is acute, find sin²A/(1 + cos²A).

Show answer and solution

A = 60°.

Expression = ( $\frac{3}{4}$ )/(1 + $\frac{1}{4}$ ).

Answer: $\frac{3}{5}$

Appeared in: 2025 · Standard · Main

Standard2 marks · internal OR

Q43.(a) In triangle ABC, right-angled at B, A(3, 4), B(1, 1) and C(−8, 7). Find tan A.

Show answer and solution

AB = √[(3 − 1)² + (4 − 1)²] = √13.

BC = √[(−8 − 1)² + (7 − 1)²] = 3√13.

tan A = BC/AB.

Answer: tan A = 3

(b) Using the distance formula, prove that A(2, 3), B(−7, 0) and C(−1, 2) are collinear.

Show answer and solution

AC = √[(−1 − 2)² + (2 − 3)²] = √10.

CB = √[(−7 + 1)² + (0 − 2)²] = 2√10.

AB = √[(−7 − 2)² + (0 − 3)²] = 3√10.

AC + CB = AB.

Answer: A, C and B are collinear

Appeared in: 2026 · Standard · Main

Mixed-chapter OR pair: the coordinate-geometry alternative is retained with its original partner.
Related: Coordinate Geometry 2-mark PYQs

Basic2 marks

Q44.If cos A = $\frac{4}{5}$ , find (sec A − tan A)/(sec A + tan A).

Show answer and solution

For acute A, sin A = $\frac{3}{5}$ .

sec A = $\frac{5}{4}$ and tan A = $\frac{3}{4}$ .

Expression = ( $\frac{5}{4}$ − $\frac{3}{4}$ )/( $\frac{5}{4}$ + $\frac{3}{4}$ ).

Answer: $\frac{1}{4}$

Appeared in: 2026 · Basic · Second board examination

Basic2 marks · internal OR

Q45.(a) If cosec A = √10, find tan A and sec A.

Show answer and solution

Take hypotenuse = √10 k and opposite = k.

Adjacent = √(10k² − k²) = 3k.

Answer: tan A = $\frac{1}{3}$ ; sec A = $\frac{\sqrt{10}}{3}$

(b) Prove: cos⁴A − sin⁴A = 2 cos²A − 1.

Show answer and solution

LHS = (cos²A − sin²A)(cos²A + sin²A).

= cos²A − sin²A.

= cos²A − (1 − cos²A).

Answer: LHS = 2 cos²A − 1 = RHS

Appeared in: 2024 · Basic · Main

Basic2 marks

Q46.If sin α = $\frac{1}{2}$ , find 3 cos α − 4 cos³α.

Show answer and solution

cos²α = 1 − sin²α = $\frac{3}{4}$ .

3 cos α − 4 cos³α = cos α(3 − 4 cos²α).

= cos α(3 − 3).

Answer: 0

Appeared in: 2023 · Basic · Main

Basic2 marks · internal OR

Q47.(a) If tan A = $\frac{8}{7}$ , find [(1 − sin A)(1 + sin A)]/[(1 − cos A)(1 + cos A)].

Show answer and solution

Expression = cos²A/sin²A = cot²A.

cot A = $\frac{7}{8}$ .

Answer: $\frac{49}{64}$

(b) In triangle ABC, right-angled at C, tan A = √3. Find sin A cos B + cos A sin B.

Show answer and solution

A = 60°, B = 30°.

Expression = ( $\frac{\sqrt{3}}{2}$ )( $\frac{\sqrt{3}}{2}$ ) + ( $\frac{1}{2}$ )( $\frac{1}{2}$ ).

Answer: 1

Appeared in: 2023 · Basic · Main

Basic2 marks · internal OR

Q48.(a) If tan A = √3, find cos²A − sin²A.

Show answer and solution

A = 60°.

cos²A − sin²A = $\frac{1}{4}$ − $\frac{3}{4}$ .

Answer: − $\frac{1}{2}$

(b) If x sin 60° + cos 30° − tan 45° = $\frac{\sqrt{3}}{2}$ , find x.

Show answer and solution

x√ $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ − 1 = $\frac{\sqrt{3}}{2}$ .

x√ $\frac{3}{2}$ = 1.

Answer: x = $\frac{2}{\sqrt{3}}$

Appeared in: 2024 · Basic · Supplementary

Standard2 marks

Q49.If 12 cosec A = 13, find (2 sin A − 3 cos A)/(4 sin A − 9 cos A).

Show answer and solution

sin A = $\frac{12}{13}$ and, for acute A, cos A = $\frac{5}{13}$ .

Expression = (24 − 15)/(48 − 45).

Answer: 3

Appeared in: 2024 · Standard · Supplementary

Basic2 marks

Q50.If 3 cot A = 4, find cos²A − sin²A.

Show answer and solution

cot A = $\frac{4}{3}$ .

Use adjacent = 4k, opposite = 3k, hypotenuse = 5k.

cos²A − sin²A = $\frac{16}{25}$ − $\frac{9}{25}$ .

Answer: $\frac{7}{25}$

Appeared in: 2025 · Basic · Supplementary

Standard2 marks · internal OR

Q51.(a) If 7 sin²A + 3 cos²A = 4, find tan A.

Show answer and solution

7 sin²A + 3(1 − sin²A) = 4.

sin²A = $\frac{1}{4}$ and cos²A = $\frac{3}{4}$ .

For acute A, tan A is positive.

Answer: tan A = $\frac{1}{\sqrt{3}}$

(b) If 4 tan A = 3, find (cosec²A + 1)/(cosec²A − 1).

Show answer and solution

tan A = $\frac{3}{4}$ , so cot²A = $\frac{16}{9}$ .

cosec²A = 1 + cot²A = $\frac{25}{9}$ .

Expression = ( $\frac{25}{9}$ + 1)/( $\frac{25}{9}$ − 1).

Answer: $\frac{17}{8}$

Appeared in: 2026 · Standard · Main

Basic2 marks

Q52.If cot A = $\frac{7}{8}$ , find [(1 + sin A)(1 − sin A)]/[(1 + cos A)(1 − cos A)].

Show answer and solution

Expression = cos²A/sin²A = cot²A.

Answer: $\frac{49}{64}$

Appeared in: 2022–23 · Basic · Official sample

Basic2 marks

Q53.If 3 sin θ = 4 cos θ, find sin θ + cos²θ − 1.

Show answer and solution

tan θ = $\frac{4}{3}$ .

For acute θ, sin θ = $\frac{4}{5}$ and cos θ = $\frac{3}{5}$ .

Expression = $\frac{4}{5}$ + $\frac{9}{25}$ − 1.

Answer: $\frac{4}{25}$

Appeared in: 2026–27 · Basic · Official sample

Standard2 marks

Q54.If tan A = $\frac{1}{\sqrt{5}}$ , find (cosec²A − sec²A)/(cosec²A + sec²A).

Show answer and solution

sec²A = 1 + $\frac{1}{5}$ = $\frac{6}{5}$ .

cosec²A = 1 + 5 = 6.

Expression = (6 − $\frac{6}{5}$ )/(6 + $\frac{6}{5}$ ).

Answer: $\frac{2}{3}$

Appeared in: 2026–27 · Standard · Official sample

Finding angles

Basic2 marks

Q55.If sin(A − B) = $\frac{1}{2}$ and tan(A + B) = √3, where 0° ≤ A + B < 90° and A > B, find A and B.

Show answer and solution

A − B = 30°, A + B = 60°.

Add: 2A = 90°.

Subtract: 2B = 30°.

Answer: A = 45°, B = 15°

Appeared in: 2026 · Basic · Main

Basic2 marks

Q56.If sin 2A = $\frac{\sqrt{3}}{2}$ and 2 tan B + 1 = 3, find A + B.

Show answer and solution

tan B = 1, so B = 45° in the acute-angle setting.

sin 2A = $\frac{\sqrt{3}}{2}$ permits 2A = 60° or 120°.

Thus A = 30° or 60°.

Answer: A + B = 75° or 105°

Appeared in: 2026 · Basic · Main

The marking scheme gives 75°. Without a restriction making 2A acute, the printed condition also permits 105°; both values are shown.

Basic2 marks

Q57.If sin A = $\frac{1}{2}$ and tan B = √3, verify cos(A + B) = cos A cos B − sin A sin B.

Show answer and solution

For acute angles, A = 30°, B = 60°.

LHS = cos 90° = 0.

RHS = ( $\frac{\sqrt{3}}{2}$ )( $\frac{1}{2}$ ) − ( $\frac{1}{2}$ )( $\frac{\sqrt{3}}{2}$ ) = 0.

Answer: LHS = RHS; verified

Appeared in: 2026 · Basic · Main

Standard2 marks · internal OR

Q58.(a) If A and B are acute angles, sec(2A − B) = √2 and cosec(A + B) = 2, find A and B.

Show answer and solution

A + B = 30° in the stated acute-angle setting.

2A − B = 45°.

Adding gives 3A = 75°.

Answer: A = 25°, B = 5°

(b) Evaluate: (2 cos 30° − cot³60°)/tan 30°.

Show answer and solution

= (√3 − 1/(3√3))/( $\frac{1}{\sqrt{3}}$ ).

= 3 − $\frac{1}{3}$ .

Answer: $\frac{8}{3}$

Appeared in: 2026 · Standard · Main

Standard2 marks · internal OR

Q59.(a) If cos(A + B) = $\frac{1}{2}$ and tan(A − B) = $\frac{1}{\sqrt{3}}$ , where A + B lies between 0° and 90°, find sec(2A − 3B).

Show answer and solution

A + B = 60°, A − B = 30°.

A = 45°, B = 15°.

2A − 3B = 45°.

Answer: sec(2A − 3B) = √2

(b) If 3 tan²60° − x sin²45° + ( $\frac{3}{4}$ )sec²30° = 2 cosec²30°, find x.

Show answer and solution

9 − x/2 + 1 = 8.

x/2 = 2.

Answer: x = 4

Appeared in: 2024 · Standard · Supplementary

Basic2 marks · internal OR

Q60.(a) If sin(A + 2B) = $\frac{\sqrt{3}}{2}$ and cos(A + 4B) = 0, where A > B, find A and B.

Show answer and solution

Using the first-quadrant angle values, A + 2B = 60° and A + 4B = 90°.

Subtract: 2B = 30°.

Then A = 60° − 30°.

Answer: A = 30°, B = 15°

(b) Prove: (1 + sin A)(1 − sin A)(1 + tan²A) = 1.

Show answer and solution

LHS = (1 − sin²A)sec²A.

= cos²A × 1/cos²A.

Answer: LHS = 1 = RHS

Appeared in: 2025 · Basic · Supplementary

Basic2 marks · internal OR

Q61.(a) If tan(A + B) = 1 and tan(A − B) = $\frac{1}{\sqrt{3}}$ , where A and B lie between 0° and 90°, find A and B.

Show answer and solution

A + B = 45°, A − B = 30°.

Add: 2A = 75°.

Subtract: 2B = 15°.

Answer: A = 37.5°, B = 7.5°

(b) Prove geometrically that tan 45° = 1.

Show answer and solution

Draw a right triangle with one acute angle 45°.

The other acute angle is 180° − 90° − 45° = 45°.

Sides opposite equal angles are equal.

tan 45° = opposite leg/adjacent leg.

Answer: tan 45° = 1

Appeared in: 2025 · Basic · Main

Basic2 marks · internal OR

Q62.(a) If sin 3A = 1, find cos 2A − tan²45°.

Show answer and solution

In the chapter’s angle setting, 3A = 90°, so A = 30°.

cos 2A − tan²45° = cos 60° − 1.

Answer: − $\frac{1}{2}$

(b) If (sec A + tan A)(1 − sin A) = k cos A, find k.

Show answer and solution

LHS = [(1 + sin A)/cos A](1 − sin A).

= (1 − sin²A)/cos A = cos A.

Hence k cos A = cos A.

Answer: k = 1

Appeared in: 2025 · Basic · Main

Standard2 marks · internal OR

Q63.(a) If sin(2A + 3B) = 1 and cos(2A − 3B) = $\frac{\sqrt{3}}{2}$ , where 0° < 2A + 3B ≤ 90° and A > B, find A and B.

Show answer and solution

2A + 3B = 90°, 2A − 3B = 30°.

Add: 4A = 120°.

Subtract: 6B = 60°.

Answer: A = 30°, B = 10°

(b) In the given figure, find sin α.

Right triangle with the top vertex a right angle, left side 6 cm, base split into 3 cm and 9 cm by a perpendicular, and angle alpha at the right-hand base vertex. — Original CBSE question figure; use the stated measurements.

Show answer and solution

The side opposite α is 6 cm.

The hypotenuse is 3 + 9 = 12 cm.

sin α = opposite/hypotenuse = $\frac{6}{12}$ .

Answer: sin α = $\frac{1}{2}$

Appeared in: 2025 · Standard · Supplementary

Standard2 marks · internal OR

Q64.(a) If sin(A + B) = 1 and cos(A − B) = $\frac{\sqrt{3}}{2}$ , where 0° < A + B ≤ 90° and A > B, find A and B.

Show answer and solution

A + B = 90°, A − B = 30°.

Add: 2A = 120°.

Answer: A = 60°, B = 30°

(b) If (cos θ − sin θ)/(cos θ + sin θ) = (1 − √3)/(1 + √3), where θ is acute, find θ.

Show answer and solution

Divide numerator and denominator on the left by cos θ.

(1 − tan θ)/(1 + tan θ) = (1 − √3)/(1 + √3).

Cross-multiply and simplify: tan θ = √3.

Answer: θ = 60°

Appeared in: 2022–23 · Standard · Official sample

Standard2 marks · internal OR

Q65.(a) If tan(A + B) = √3 and tan(A − B) = $\frac{1}{\sqrt{3}}$ , where 0° < A + B < 90° and A > B, find A and B.

Show answer and solution

A + B = 60°, A − B = 30°.

Add: 2A = 90°.

Answer: A = 45°, B = 15°

(b) If 2 cosec²30° + x sin²60° − ( $\frac{3}{4}$ )tan²30° = 10, find x.

Show answer and solution

8 + 3x/4 − $\frac{1}{4}$ = 10.

3x/4 = $\frac{9}{4}$ .

Answer: x = 3

Appeared in: 2023–24 · Standard · Official sample

Basic2 marks

Q66.If sin(A − B) = $\frac{1}{2}$ and cos(A + B) = $\frac{1}{2}$ , where 0° < A + B < 90° and A > B, find A and B.

Show answer and solution

A − B = 30°, A + B = 60°.

Add: 2A = 90°.

Answer: A = 45°, B = 15°

Appeared in: 2024–25 · Basic · Official sample

Standard2 marks

Q67.If sin(A + B) = 1 and cos(A − B) = $\frac{\sqrt{3}}{2}$ , where A and B are acute angles, find A and B.

Show answer and solution

A + B = 90°.

Since −90° < A − B < 90°, A − B = 30° or −30°.

Solve the two pairs separately.

Answer: (A, B) = (60°, 30°) or (30°, 60°)

Appeared in: 2025–26 · Standard · Official sample

The sample marking scheme gives A = 60°, B = 30°. The printed question does not state A > B, so the reversed order also satisfies both conditions.

Identities and algebraic relations

Standard2 marks

Q68.If 1 + sin²θ = 3 sin θ cos θ, find tan θ.

Show answer and solution

cos θ cannot be zero, since that would give 2 = 0.

Divide by cos²θ: sec²θ + tan²θ = 3 tan θ.

Put t = tan θ: 2t² − 3t + 1 = 0.

(2t − 1)(t − 1) = 0.

Answer: tan θ = 1 or $\frac{1}{2}$

Appeared in: 2025 · Standard · Supplementary

Standard2 marks · internal OR

Q69.(a) If (sin θ + cosec θ)² + (cos θ + sec θ)² = x + tan²θ + cot²θ, find x.

Show answer and solution

LHS = sin²θ + cos²θ + 4 + cosec²θ + sec²θ.

= 1 + 4 + (1 + cot²θ) + (1 + tan²θ).

= 7 + tan²θ + cot²θ.

Answer: x = 7

(b) Evaluate: (3 sin 30° − 4 sin³30°)/(2 sin²50° + 2 cos²50°).

Show answer and solution

Numerator = $\frac{3}{2}$ − $\frac{4}{8}$ = 1.

Denominator = 2(sin²50° + cos²50°) = 2.

Answer: $\frac{1}{2}$

Appeared in: 2025 · Standard · Main

Standard2 marks · internal OR

Q70.(a) If a sec θ + b tan θ = m and b sec θ + a tan θ = n, prove a² + n² = b² + m².

Show answer and solution

m² − n² = (a sec θ + b tan θ)² − (b sec θ + a tan θ)².

= (a² − b²)(sec²θ − tan²θ).

= a² − b².

Rearrange.

Answer: a² + n² = b² + m²

(b) Derive 1 + tan²A = sec²A from sin²A + cos²A = 1. Hence find tan A if sec A = $\frac{5}{3}$ .

Show answer and solution

Divide sin²A + cos²A = 1 by cos²A.

tan²A + 1 = sec²A.

tan²A = $\frac{25}{9}$ − 1 = $\frac{16}{9}$ .

For acute A, tan A is positive.

Answer: 1 + tan²A = sec²A; tan A = $\frac{4}{3}$

Appeared in: 2025 · Standard · Main

Standard2 marks · internal OR

Q71.(a) Prove: (1 − cos A)/(1 + cos A) = (cot A − cosec A)².

Show answer and solution

RHS = [(cos A − 1)/sin A]².

= (1 − cos A)²/(1 − cos²A).

= (1 − cos A)/(1 + cos A).

Answer: RHS = LHS

(b) Prove: tan A/(1 − cot A) + cot A/(1 − tan A) = 1 + sec A cosec A.

Show answer and solution

Put t = tan A, so cot A = 1/t; the printed denominators require t ≠ 1.

LHS = t²/(t − 1) − 1/[t(t − 1)].

= (t³ − 1)/[t(t − 1)] = (t² + t + 1)/t.

= 1 + tan A + cot A.

tan A + cot A = 1/(sin A cos A) = sec A cosec A.

Answer: LHS = 1 + sec A cosec A = RHS

Appeared in: 2023 · Standard · Supplementary

Standard2 marks · internal OR

Q72.(a) If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove a² + b² = m² + n².

Show answer and solution

m² + n² = (a cos θ + b sin θ)² + (a sin θ − b cos θ)².

The cross terms cancel.

= a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ).

Answer: m² + n² = a² + b²

(b) Prove: √[(sec A − 1)/(sec A + 1)] + √[(sec A + 1)/(sec A − 1)] = 2 cosec A.

Show answer and solution

For acute A, sin A and 1 ± cos A are positive.

First root = √[(1 − cos A)/(1 + cos A)] = (1 − cos A)/sin A.

Second root = (1 + cos A)/sin A.

Their sum = 2/sin A.

Answer: LHS = 2 cosec A = RHS

Appeared in: 2023 · Standard · Main

Basic2 marks · internal OR

Q73.(a) Prove: 1 − sin²A/(1 + cos A) = cos A.

Show answer and solution

LHS = 1 − (1 − cos²A)/(1 + cos A).

= 1 − (1 − cos A).

Answer: LHS = cos A = RHS

(b) Prove: (1 − cot²θ)/(1 + cot²θ) + 2 cos²θ = 1.

Show answer and solution

First fraction = (sin²θ − cos²θ)/(sin²θ + cos²θ).

LHS = sin²θ − cos²θ + 2 cos²θ.

= sin²θ + cos²θ.

Answer: LHS = 1 = RHS

Appeared in: 2023 · Basic · Supplementary

Standard2 marks · internal OR

Q74.(a) Prove: (cosec A + sin A)(cosec A − sin A) = cot²A + cos²A.

Show answer and solution

LHS = cosec²A − sin²A.

= 1 + cot²A − sin²A.

= cot²A + cos²A.

Answer: LHS = RHS

(b) Evaluate: (5 tan²30° + 3 cos²45° − 4 sin²30°)/(√3 sin 60° cos 60° + cot²45°).

Show answer and solution

Numerator = $\frac{5}{3}$ + $\frac{3}{2}$ − 1 = $\frac{13}{6}$ .

Denominator = √3( $\frac{\sqrt{3}}{2}$ )( $\frac{1}{2}$ ) + 1 = $\frac{7}{4}$ .

Expression = ( $\frac{13}{6}$ )/( $\frac{7}{4}$ ).

Answer: $\frac{26}{21}$

Appeared in: 2025 · Standard · Main

Basic2 marks · internal OR

Q75.(a) Prove: sin A/(1 + cos A) + (1 + cos A)/sin A = 2 cosec A.

Show answer and solution

LHS = [sin²A + (1 + cos A)²]/[sin A(1 + cos A)].

= [2(1 + cos A)]/[sin A(1 + cos A)].

= 2/sin A.

Answer: LHS = 2 cosec A = RHS

(b) If sin(A + 2B) = $\frac{\sqrt{3}}{2}$ and cos(A + 4B) = 0, where A > B, find A and B.

Show answer and solution

Using the chapter’s first-quadrant angle values: A + 2B = 60°, A + 4B = 90°.

Subtract: 2B = 30°.

Then A = 30°.

Answer: A = 30°, B = 15°

Appeared in: 2026 · Basic · Main

Basic2 marks

Q76.Prove: (1 − cos A)/(1 + cos A) = (cosec A − cot A)².

Show answer and solution

RHS = [(1 − cos A)/sin A]².

= (1 − cos A)²/[(1 − cos A)(1 + cos A)].

Answer: RHS = (1 − cos A)/(1 + cos A) = LHS

Appeared in: 2026 · Basic · Second board examination

Standard2 marks · internal OR

Q77.(a) If sin θ + sin²θ = 1, find cos⁴θ + cos²θ.

Show answer and solution

cos²θ = 1 − sin²θ = sin θ.

cos⁴θ + cos²θ = sin²θ + sin θ.

Answer: 1

(b) If tan A = $\frac{1}{\sqrt{7}}$ , find (cosec²A − sec²A)/(cosec²A + sec²A).

Show answer and solution

sec²A = 1 + $\frac{1}{7}$ = $\frac{8}{7}$ .

cosec²A = 1 + 7 = 8.

Expression = (8 − $\frac{8}{7}$ )/(8 + $\frac{8}{7}$ ).

Answer: $\frac{3}{4}$

Appeared in: 2023 · Standard · Main

Standard2 marks · internal OR

Q78.(a) If tan θ + sec θ = m, prove sec θ = (m² + 1)/(2m).

Show answer and solution

(sec θ + tan θ)(sec θ − tan θ) = 1, so m ≠ 0.

sec θ − tan θ = 1/m.

Add to sec θ + tan θ = m.

2 sec θ = m + 1/m.

Answer: sec θ = (m² + 1)/(2m)

(b) If sin A = $\frac{3}{5}$ and cos B = $\frac{12}{13}$ , find tan A + tan B.

Show answer and solution

For acute A and B, cos A = $\frac{4}{5}$ and sin B = $\frac{5}{13}$ .

tan A = $\frac{3}{4}$ , tan B = $\frac{5}{12}$ .

Their sum = $\frac{9}{12}$ + $\frac{5}{12}$ .

Answer: $\frac{7}{6}$

Appeared in: 2024 · Standard · Main

Standard2 marks

Q79.If tan A + cot A = 6, find tan²A + cot²A − 4.

Show answer and solution

Square the given equation.

tan²A + cot²A + 2 tan A cot A = 36.

tan A cot A = 1, so tan²A + cot²A = 34.

Answer: 30

Appeared in: 2025 · Standard · Main

Basic2 marks

Q80.Prove: (a sin θ + b cos θ)² + (a cos θ − b sin θ)² = a² + b².

Show answer and solution

Expand the two squares; the cross terms cancel.

LHS = a²(sin²θ + cos²θ) + b²(cos²θ + sin²θ).

Answer: LHS = a² + b² = RHS

Appeared in: 2026 · Basic · Second board examination

Basic2 marks

Q81.Prove: (sec A − cosec A)(1 + tan A + cot A) = tan A sec A − cot A cosec A.

Show answer and solution

Expand LHS: sec A − cosec A + sec A tan A − cosec A tan A + sec A cot A − cosec A cot A.

Use cosec A tan A = sec A and sec A cot A = cosec A.

The first, second, fourth and fifth terms cancel.

Answer: LHS = tan A sec A − cot A cosec A = RHS

Appeared in: 2026 · Basic · Second board examination

Standard2 marks · internal OR

Q82.(a) Prove: √[(1 + sin A)/(1 − sin A)] = sec A + tan A.

Show answer and solution

For acute A, cos A is positive.

LHS = √[(1 + sin A)²/(1 − sin²A)].

= (1 + sin A)/cos A.

Answer: LHS = sec A + tan A = RHS

(b) Evaluate: (3 cos²30° − 6 cosec²30°)/tan²60°.

Show answer and solution

= ( $\frac{9}{4}$ − 24)/3.

= (− $\frac{87}{4}$ )/3.

Answer: − $\frac{29}{4}$

Appeared in: 2026 · Standard · Main

Standard2 marks · internal OR

Q83.(a) Prove: tan A/(1 + tan²A) + cot A/(1 + cot²A) = 2 sin A cos A.

Show answer and solution

LHS = tan A/sec²A + cot A/cosec²A.

= (sin A/cos A)cos²A + (cos A/sin A)sin²A.

= sin A cos A + sin A cos A.

Answer: LHS = 2 sin A cos A = RHS

(b) Evaluate: (1 − 2 tan²30° − sec²45°)/sin²60°.

Show answer and solution

= (1 − $\frac{2}{3}$ − 2)/( $\frac{3}{4}$ ).

= (− $\frac{5}{3}$ )( $\frac{4}{3}$ ).

Answer: − $\frac{20}{9}$

Appeared in: 2026 · Standard · Main

Standard2 marks · internal OR

Q84.(a) If tan θ + 1/tan θ = 2, find tan²θ + 1/tan²θ.

Show answer and solution

Square the given equation.

tan²θ + 2 + 1/tan²θ = 4.

Answer: 2

(b) Prove: √[(1 − sin A)/(1 + sin A)] = sec A − tan A.

Show answer and solution

For acute A, 1 − sin A and cos A are positive.

LHS = √[(1 − sin A)²/(1 − sin²A)].

= (1 − sin A)/cos A.

Answer: LHS = sec A − tan A = RHS

Appeared in: 2026 · Standard · Main

Trigonometry questions students ask

Which identities are useful for two-mark trigonometry questions?

Use sin²A + cos²A = 1, sec²A − tan²A = 1 and cosec²A − cot²A = 1. Choose the identity that connects the given ratio with the required one. Expand a square only when it helps reach the required expression.

How do I find the remaining ratios from tan A?

Use proportional lengths for the opposite and adjacent sides, then apply Pythagoras to find the hypotenuse. For example, tan A = $\frac{24}{7}$ gives side lengths proportional to 24, 7 and 25, so sin A = $\frac{24}{25}$ and cos A = $\frac{7}{25}$ .

How should I begin an identity proof?

Begin with the more complicated side. Use reciprocal relations, a common denominator or a standard identity to simplify it into the other side. Write one operation per line and keep the proof as a connected chain of equal expressions.

Do I need addition formulas for this chapter’s two-mark questions?

When a question supplies an addition or subtraction formula, use the given formula. For a numerical verification, substitute the stated angles into each side separately and compare the values. An advanced general proof is not needed for those verification questions.

Is sin²A the same as sin 2A?

No. sin²A means (sin A)², while sin 2A means the sine of twice the angle. At A = 30°, sin²A = $\frac{1}{4}$ but sin 2A = sin 60° = $\frac{\sqrt{3}}{2}$ .

Can an angle equation have two valid answers?

Yes. For example, sin 2A = $\frac{\sqrt{3}}{2}$ permits 2A = 60° or 120° when the stated interval permits both. Check the given angle restrictions before rejecting a value. The two possibilities give A = 30° or 60°.

Why is the positive square root used for cos A?

For an acute angle, cos A is positive, so cos A = √(1 − sin²A). If a different angle interval is stated, its sign must be checked before taking the square root. The acute-angle condition matters in square-root identity proofs too.

Should I solve both OR alternatives in the examination?

Answer one alternative from an OR pair, following the paper’s instructions. Both alternatives are solved here for practice. They remain together as one original question group, with a separate solution and answer-summary row for each alternative.

How to write a two-mark solution

State the identity, standard value or side relationship you need.
Substitute and simplify, with one operation on each line.
Write the final value, ratio or proved equality separately.

For related practice, use Triangles 2-mark PYQs for right-triangle reasoning, and Some Applications of Trigonometry for heights and distances.

Answer summary · 130 worked answers

Rows follow the selected filters. Open a question link to see its solution.

Question	Answer
Q1	12
Q2	(√6 + √2)/4
Q3 (a)	$\frac{7}{4}$
Q3 (b)	LHS = RHS; verified
Q4 (a)	$\frac{67}{12}$
Q4 (b)	A = 30°, B = 30°
Q5 (a)	9
Q5 (b)	$\frac{1}{2}$
Q6	− $\frac{17}{3}$
Q7	6
Q8	− $\frac{3}{2}$
Q9 (a)	LHS = RHS; verified
Q9 (b)	$\frac{7}{4}$
Q10	11
Q11	− $\frac{7}{2}$
Q12 (a)	x + 2y = 11
Q12 (b)	2
Q13 (a)	$\frac{9}{16}$
Q13 (b)	LHS = RHS; verified
Q14	(√6 + √2)/4
Q15 (a)	p = 0
Q15 (b)	1
Q16 (a)	(6√3 − 9)/2
Q16 (b)	LHS = RHS; verified
Q17	0
Q18 (a)	4
Q18 (b)	A = 30°, B = 30°
Q19 (a)	$\frac{67}{12}$
Q19 (b)	A = 45°, B = 15°
Q20 (a)	$\frac{\sqrt{6}}{5}$
Q20 (b)	LHS = RHS; verified
Q21 (a)	4
Q21 (b)	LHS = RHS; verified
Q22 (a)	$\frac{67}{12}$
Q22 (b)	LHS = 2 sec A = RHS
Q23	0
Q24	− $\frac{1}{4}$
Q25	x = 1
Q26	(√6 + 3)/[4(1 + √3)]
Q27	15
Q28	16√ $\frac{3}{3}$
Q29	0
Q30 (a)	x = 3
Q30 (b)	25
Q31	LHS = RHS; verified
Q32 (a)	$\frac{67}{12}$
Q32 (b)	LHS = cosec α = RHS
Q33	2
Q34	$\frac{7}{12}$
Q35 (a)	√3 − 1
Q35 (b)	1
Q36	$\frac{1}{4}$
Q37 (a)	cosec A = $\frac{25}{24}$ ; tan A = $\frac{24}{7}$
Q37 (b)	LHS = RHS; verified
Q38 (a)	(√6 − √2)/4
Q38 (b)	cos A = √(1 − y²); tan A = y/√(1 − y²)
Q39 (a)	$\frac{\sqrt{2}}{4}$
Q39 (b)	$\frac{169}{60}$
Q40 (a)	$\frac{31}{25}$
Q40 (b)	$\frac{49}{64}$
Q41 (a)	$\frac{1}{\sqrt{2}}$
Q41 (b)	$\frac{15}{2}$
Q42	$\frac{3}{5}$
Q43 (a)	tan A = 3
Q43 (b)	A, C and B are collinear
Q44	$\frac{1}{4}$
Q45 (a)	tan A = $\frac{1}{3}$ ; sec A = $\frac{\sqrt{10}}{3}$
Q45 (b)	LHS = 2 cos²A − 1 = RHS
Q46	0
Q47 (a)	$\frac{49}{64}$
Q47 (b)	1
Q48 (a)	− $\frac{1}{2}$
Q48 (b)	x = $\frac{2}{\sqrt{3}}$
Q49	3
Q50	$\frac{7}{25}$
Q51 (a)	tan A = $\frac{1}{\sqrt{3}}$
Q51 (b)	$\frac{17}{8}$
Q52	$\frac{49}{64}$
Q53	$\frac{4}{25}$
Q54	$\frac{2}{3}$
Q55	A = 45°, B = 15°
Q56	A + B = 75° or 105°
Q57	LHS = RHS; verified
Q58 (a)	A = 25°, B = 5°
Q58 (b)	$\frac{8}{3}$
Q59 (a)	sec(2A − 3B) = √2
Q59 (b)	x = 4
Q60 (a)	A = 30°, B = 15°
Q60 (b)	LHS = 1 = RHS
Q61 (a)	A = 37.5°, B = 7.5°
Q61 (b)	tan 45° = 1
Q62 (a)	− $\frac{1}{2}$
Q62 (b)	k = 1
Q63 (a)	A = 30°, B = 10°
Q63 (b)	sin α = $\frac{1}{2}$
Q64 (a)	A = 60°, B = 30°
Q64 (b)	θ = 60°
Q65 (a)	A = 45°, B = 15°
Q65 (b)	x = 3
Q66	A = 45°, B = 15°
Q67	(A, B) = (60°, 30°) or (30°, 60°)
Q68	tan θ = 1 or $\frac{1}{2}$
Q69 (a)	x = 7
Q69 (b)	$\frac{1}{2}$
Q70 (a)	a² + n² = b² + m²
Q70 (b)	1 + tan²A = sec²A; tan A = $\frac{4}{3}$
Q71 (a)	RHS = LHS
Q71 (b)	LHS = 1 + sec A cosec A = RHS
Q72 (a)	m² + n² = a² + b²
Q72 (b)	LHS = 2 cosec A = RHS
Q73 (a)	LHS = cos A = RHS
Q73 (b)	LHS = 1 = RHS
Q74 (a)	LHS = RHS
Q74 (b)	$\frac{26}{21}$
Q75 (a)	LHS = 2 cosec A = RHS
Q75 (b)	A = 30°, B = 15°
Q76	RHS = (1 − cos A)/(1 + cos A) = LHS
Q77 (a)	1
Q77 (b)	$\frac{3}{4}$
Q78 (a)	sec θ = (m² + 1)/(2m)
Q78 (b)	$\frac{7}{6}$
Q79	30
Q80	LHS = a² + b² = RHS
Q81	LHS = tan A sec A − cot A cosec A = RHS
Q82 (a)	LHS = sec A + tan A = RHS
Q82 (b)	− $\frac{29}{4}$
Q83 (a)	LHS = 2 sin A cos A = RHS
Q83 (b)	− $\frac{20}{9}$
Q84 (a)	2
Q84 (b)	LHS = sec A − tan A = RHS

About these questions and solutions

Solutions prepared by Team TutorVertex in NCERT-style steps. Repeated full groups are merged; original OR pairs stay together.
Some subparts recur across pairs or printed variants. The 84 groups are not 84 unique individual prompts.

Official sources and question details

PDF pages below are counted from the first page of each file. Paper labels under the questions remain plain text.

Question 1 sources

2023 · Basic · 430/6/3 · Q23
Question paper: MATHEMATICS_BASIC/430_6_3_Maths Basic.pdf · PDF p. 13.
Marking scheme: Math_Basic/430-6-3.pdf · PDF p. 8.

Question 2 sources

2024 · Basic · 430/1/1 · Q24
Question paper: MATHEMATICS_BASIC/430-1-1MATHEMATICS (BASIC).pdf · PDF p. 9.
Marking scheme: Mathematics/X_241_MS_430_1_1_Unsigned.pdf · PDF p. 8.

2024 · Basic · 430/1/3 · Q24
Question paper: MATHEMATICS_BASIC/430-1-3MATHEMATICS (BASIC).pdf · PDF p. 9.
Marking scheme: Mathematics/X_241_MS_430_1_3_Unsigned.pdf · PDF p. 8.

Question 3 sources

2025 · Basic · 430/5/1 · Q22
Question paper: 241_Mathematics_Basic/430-5-1_Mathematics Basic.pdf · PDF p. 13.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_1.pdf · PDF p. 7.

2025 · Basic · 430/5/2 · Q24
Question paper: 241_Mathematics_Basic/430-5-2_Mathematics Basic.pdf · PDF p. 13.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_2.pdf · PDF p. 7.

Question 4 sources

2023 · Standard · 30/4/1 · Q25
Question paper: MATHEMATICS_STANDARD/30_4_1 Maths Standard.pdf · PDF p. 9.
Marking scheme: Maths_Standard/MS 041_30-4-1 Mathematics 2022-23.pdf · PDF p. 9.

Question 5 sources

2023 · Standard · 30/4/3 · Q22
Question paper: MATHEMATICS_STANDARD/30_4_3 Maths Standard.pdf · PDF p. 9.
Marking scheme: Maths_Standard/MS 041_30-4-3 Mathematics 2022-23.pdf · PDF p. 8.

Question 6 sources

2023 · Basic · 430/1/1 · Q22
Question paper: MATHEMATICS_BASIC/430_1_1_Maths Basic.pdf · PDF p. 13.
Marking scheme: Math_Basic/430_1_1.pdf · PDF p. 8.

Question 7 sources

2023 · Basic · 430/1/2 · Q21
Question paper: MATHEMATICS_BASIC/430_1_2_Maths Basic.pdf · PDF p. 13.
Marking scheme: Math_Basic/430_1_2.pdf · PDF p. 7.

Question 8 sources

2023 · Basic · 430/1/3 · Q21
Question paper: MATHEMATICS_BASIC/430_1_3_Maths Basic.pdf · PDF p. 13.
Marking scheme: Math_Basic/430_1_3.pdf · PDF p. 7.

Question 9 sources

2023 · Basic · 430/C/1 · Q24
Question paper: Maths_Basic/430_C_1 Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Mathematics_Basic/241_C_1.pdf · PDF p. 8.

2023 · Basic · 430/C/2 · Q22
Question paper: Maths_Basic/430_C_2 Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Mathematics_Basic/241_C_2.pdf · PDF p. 8.

2023 · Basic · 430/C/3 · Q25
Question paper: Maths_Basic/430_C_3 Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Mathematics_Basic/241_C_3.pdf · PDF p. 9.

Question 10 sources

2024 · Basic · 430/3/2 · Q25
Question paper: MATHEMATICS_BASIC/430-3-2MATHEMATICS (BASIC).pdf · PDF p. 15.
Marking scheme: Mathematics/X_241_MS_430_3_2_Unsigned.pdf · PDF p. 10.

Question 11 sources

2024 · Basic · 430/3/3 · Q21
Question paper: MATHEMATICS_BASIC/430-3-3MATHEMATICS (BASIC).pdf · PDF p. 13.
Marking scheme: Mathematics/X_241_MS_430_3_3_Unsigned.pdf · PDF p. 7.

Question 12 sources

2025 · Standard · 30/1/1 · Q21
Question paper: 041_Mathematics_Standard/30-1-1_Mathematics Standard.pdf · PDF p. 15.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-1-1_2024-25.pdf · PDF p. 7.

2025 · Standard · 30/1/2 · Q24
Question paper: 041_Mathematics_Standard/30-1-2_Mathematics Standard.pdf · PDF p. 15.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-1-2_2024-25.pdf · PDF p. 8.

2025 · Standard · 30/1/3 · Q22
Question paper: 041_Mathematics_Standard/30-1-3_Mathematics Standard.pdf · PDF p. 15.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-1-3_2024-25.pdf · PDF p. 7.

Question 13 sources

2026 · Basic · 430/5/2 · Q23
Question paper: Mathematics_Standard/1159-2_430-5-2 Mathematics (Basic).pdf · PDF p. 13.
Marking scheme: 241_MATHEMATICS BASIC\X_241_MS_ 430_5_2_unsigned.pdf · PDF p. 8.

Question 14 sources

2023 · Standard · 30/C/B · Q24
Question paper: Maths_Standard/30_C_B Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Mathematics_Standard/041_30-C-B.pdf · PDF p. 9.

Question 15 sources

2023 · Standard · 30/2/1 · Q22
Question paper: MATHEMATICS_STANDARD/30_2_1_Maths Standard.pdf · PDF p. 13.
Marking scheme: Maths_Standard/MS 041_30-2-1 Mathematics 2022-23.pdf · PDF p. 8.

2023 · Standard · 30/2/2 · Q23
Question paper: MATHEMATICS_STANDARD/30_2_2_Maths Standard.pdf · PDF p. 13.
Marking scheme: Maths_Standard/MS 041_30-2-2 Mathematics 2022-23.pdf · PDF p. 8.

2023 · Standard · 30/2/3 · Q24
Question paper: MATHEMATICS_STANDARD/30_2_3_Maths Standard.pdf · PDF p. 13.
Marking scheme: Maths_Standard/MS 041_30-2-3 Mathematics 2022-23.pdf · PDF p. 8.

Question 16 sources

2023 · Basic · 430/5/1 · Q22
Question paper: MATHEMATICS_BASIC/430_5_1_Maths Basic.pdf · PDF p. 11.
Marking scheme: Math_Basic/430_5_1.pdf · PDF p. 7.

2023 · Basic · 430/5/2 · Q23
Question paper: MATHEMATICS_BASIC/430_5_2_Maths Basic.pdf · PDF p. 11.
Marking scheme: Math_Basic/430_5_2.pdf · PDF p. 8.

2023 · Basic · 430/5/3 · Q23
Question paper: MATHEMATICS_BASIC/430_5_3_Maths Basic.pdf · PDF p. 11.
Marking scheme: Math_Basic/430_5_3.pdf · PDF p. 8.

Question 17 sources

2023 · Basic · 430/C/B · Q25
Question paper: Maths_Basic/430_C_B Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Mathematics_Basic/241_C_B.pdf · PDF p. 6.

Question 18 sources

2024 · Standard · 30/2/1 · Q23
Question paper: MATHEMATICS_STANDARD/30-2-1(Mathematics Standard).pdf · PDF p. 15.
Marking scheme: Mathematics_Standard/MS 041_30-2-1 Mathematics 2023-24.pdf · PDF p. 9.

2024 · Standard · 30/2/2 · Q25
Question paper: MATHEMATICS_STANDARD/30-2-2(Mathematics Standard).pdf · PDF p. 15.
Marking scheme: Mathematics_Standard/MS 041_30-2-2 Mathematics 2023-24.pdf · PDF p. 10.

2024 · Standard · 30/2/3 · Q24
Question paper: MATHEMATICS_STANDARD/30-2-3(Mathematics Standard).pdf · PDF p. 15.
Marking scheme: Mathematics_Standard/MS 041_30-2-3 Mathematics 2023-24.pdf · PDF p. 9.

Question 19 sources

2024 · Standard · 30/4/1 · Q23
Question paper: MATHEMATICS_STANDARD/30-4-1(Mathematics Standard).pdf · PDF p. 11.
Marking scheme: Mathematics_Standard/Marking scheme 30 4 1.pdf · PDF p. 7.

2024 · Standard · 30/4/2 · Q21
Question paper: MATHEMATICS_STANDARD/30-4-2(Mathematics Standard).pdf · PDF p. 11.
Marking scheme: Mathematics_Standard/Marking scheme 30 4 2.pdf · PDF p. 7.

2024 · Standard · 30/4/3 · Q22
Question paper: MATHEMATICS_STANDARD/30-4-3(Mathematics Standard).pdf · PDF p. 13.
Marking scheme: Mathematics_Standard/Marking scheme 30 4 3.pdf · PDF p. 8.

Question 20 sources

2025 · Basic · 430/6/1 · Q24
Question paper: 241_Mathematics_Basic/430-6-1_Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_6_1.pdf · PDF p. 9.

2025 · Basic · 430/6/2 · Q21
Question paper: 241_Mathematics_Basic/430-6-2_Mathematics Basic.pdf · PDF p. 13.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_6_2.pdf · PDF p. 9.

Question 21 sources

2024 · Standard · 30/1/1 · Q23
Question paper: MATHEMATICS_STANDARD/30-1-1(Mathematics Standard).pdf · PDF p. 9.
Marking scheme: Mathematics_Standard/MS 041_30-1-1 Mathematics 2023-24.pdf · PDF p. 7.

2024 · Standard · 30/1/3 · Q22
Question paper: MATHEMATICS_STANDARD/30-1-3(Mathematics Standard).pdf · PDF p. 9.
Marking scheme: Mathematics_Standard/MS 041_30-1-3 Mathematics 2023-24.pdf · PDF p. 7.

Question 22 sources

2023 · Standard · 30/B · Q24
Question paper: MATHEMATICS_STANDARD/30_B_5_Maths Std. for VI candidates.pdf · PDF p. 15.
Marking scheme: Maths_Standard/MS 041_30(B) Mathematics 2022-23.pdf · PDF p. 10.

Question 23 sources

2024 · Basic · 430/4/1 · Q25
Question paper: MATHEMATICS_BASIC/430-4-1MATHEMATICS (BASIC).pdf · PDF p. 13.
Marking scheme: Mathematics/X_241_MS_430_4_1_Unsigned.pdf · PDF p. 9.

2024 · Basic · 430/4/2 · Q22
Question paper: MATHEMATICS_BASIC/430-4-2MATHEMATICS (BASIC).pdf · PDF p. 13.
Marking scheme: Mathematics/X_241_MS_430_4_2_Unsigned.pdf · PDF p. 9.

Question 24 sources

2024 · Basic · 430/5/1 · Q21
Question paper: MATHEMATICS_BASIC/430-5-1MATHEMATICS (BASIC).pdf · PDF p. 13.
Marking scheme: Mathematics/X_241_MS_430_5_1_Unsigned.pdf · PDF p. 7.

Question 25 sources

2024 · Basic · 430/5/3 · Q24
Question paper: MATHEMATICS_BASIC/430-5-3MATHEMATICS (BASIC).pdf · PDF p. 13.
Marking scheme: Mathematics/X_241_MS_430_5_3_Unsigned.pdf · PDF p. 9.

Question 26 sources

2024 · Standard · 30/3/1 · Q25
Question paper: MATHEMATICS_STANDARD/30-3-1(Mathematics Standard).pdf · PDF p. 15.
Marking scheme: Mathematics_Standard/MS 041_30-3-1 Mathematics 2023-24.pdf · PDF p. 10.

Question 27 sources

2024 · Standard · 30/3/3 · Q22
Question paper: MATHEMATICS_STANDARD/30-3-3(Mathematics Standard).pdf · PDF p. 15.
Marking scheme: Mathematics_Standard/MS 041_30-3-3 Mathematics 2023-24.pdf · PDF p. 9.

Question 28 sources

2024 · Standard · 30/5/1 · Q22
Question paper: MATHEMATICS_STANDARD/30-5-1(Mathematics Standard).pdf · PDF p. 13.
Marking scheme: Mathematics_Standard/MS 041_30-5-1 Mathematics 2023-24.pdf · PDF p. 8.

2024 · Standard · 30/5/2 · Q23
Question paper: MATHEMATICS_STANDARD/30-5-2(Mathematics Standard).pdf · PDF p. 13.
Marking scheme: Mathematics_Standard/MS 041_30-5-2 Mathematics 2023-24.pdf · PDF p. 8.

2024 · Standard · 30/5/3 · Q25
Question paper: MATHEMATICS_STANDARD/30-5-3(Mathematics Standard).pdf · PDF p. 13.
Marking scheme: Mathematics_Standard/MS 041_30-5-3 Mathematics 2023-24.pdf · PDF p. 8.

Question 29 sources

2026 · Standard · 30/B · Q23
Question paper: Mathmatics_Std/30-B_Maths Std_E2_10_2026.pdf · PDF p. 13.
Marking scheme: Mathematics_Standard/MS_X_041_Mathematics_30B_II_Board_2025-26.pdf · PDF p. 9.

Question 30 sources

2026 · Standard · 30/7/3 · Q25
Question paper: Mathmatics_Std/30-7-3_Maths Std_E2_10_2026.pdf · PDF p. 17.
Marking scheme: Mathematics_Standard/Mathematics Standard-041 E2 (Set 30-7-3).pdf · PDF p. 11.

Question 31 sources

2026 · Basic · 430/7/2 · Q25
Question paper: Mathmatics_Basic/430-7-2_Maths Basic_E2_10_2026.pdf · PDF p. 17.
Marking scheme: Mathematics_Basic/Mathematics Basic-241 E2 (430-7-2).pdf · PDF p. 10.

Question 32 sources

2026 · Standard · 30/2/2 · Q22
Question paper: Mathematics_Standard/30_2_2_Mathematics Std._cl.10.pdf · PDF p. 17.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-2-2_2025-26 (1).pdf · PDF p. 9.

Question 33 sources

2023–24 · Basic · Basic · Q24
Question paper: MathsBasic-SQP.pdf · PDF p. 4.
Marking scheme: MathsBasic-MS.pdf · PDF p. 3.

Question 34 sources

2024–25 · Standard · Standard · Q23
Question paper: MathsStandard-SQP.pdf · PDF p. 4.
Marking scheme: MathsStandard-MS.pdf · PDF p. 2.

Question 35 sources

2025–26 · Basic · Basic · Q25
Question paper: MathsBasic-SQP.pdf · PDF p. 6.
Marking scheme: MathsBasic-MS.pdf · PDF p. 3.

Question 36 sources

2024 · Basic · 430/1/2 · Q23
Question paper: MATHEMATICS_BASIC/430-1-2MATHEMATICS (BASIC).pdf · PDF p. 9.
Marking scheme: Mathematics/X_241_MS_430_1_2_Unsigned.pdf · PDF p. 7.

Question 37 sources

2025 · Basic · 430/5/3 · Q25
Question paper: 241_Mathematics_Basic/430-5-3_Mathematics Basic.pdf · PDF p. 13.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_3.pdf · PDF p. 8.

Question 38 sources

2025 · Standard · 30/5/1 · Q23
Question paper: 041_Mathematics_Standard/30-5-1_Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-5-1_2024-25.pdf · PDF p. 7.

2025 · Standard · 30/5/2 · Q25
Question paper: 041_Mathematics_Standard/30-5-2_Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-5-2_2024-25.pdf · PDF p. 7.

2025 · Standard · 30/5/3 · Q22
Question paper: 041_Mathematics_Standard/30-5-3_Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-5-3_2024-25.pdf · PDF p. 7.

Question 39 sources

2026 · Standard · 30/7/1 · Q24
Question paper: Mathmatics_Std/30-7-1_Maths Std_E2_10_2026.pdf · PDF p. 15.
Marking scheme: Mathematics_Standard/Mathematics Standard-041 E2 (Set 30-7-1).pdf · PDF p. 10.

2026 · Standard · 30/7/2 · Q22
Question paper: Mathmatics_Std/30-7-2_Maths Std_E2_10_2026.pdf · PDF p. 15.
Marking scheme: Mathematics_Standard/Mathematics Standard-041 E2 (Set 30-7-2).pdf · PDF p. 11.

Question 40 sources

2026 · Standard · 30/1/1 · Q24
Question paper: Mathematics_Standard/30-1-1P-Maths (Std)_X.pdf · PDF p. 9.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-1-1_2025-26 (1).pdf · PDF p. 8.

2026 · Standard · 30/1/2 · Q23
Question paper: Mathematics_Standard/30-1-2P-Maths (Std)_X.pdf · PDF p. 9.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-1-2_2025-26 (1).pdf · PDF p. 9.

2026 · Standard · 30/1/3 · Q22
Question paper: Mathematics_Standard/30-1-3P-Maths (Std)_X.pdf · PDF p. 7.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-1-3_2025-26 (1).pdf · PDF p. 8.

Question 41 sources

2026 · Basic · 430/5/3 · Q24
Question paper: Mathematics_Standard/1159-3_430-5-3 Mathematics (Basic).pdf · PDF p. 15.
Marking scheme: 241_MATHEMATICS BASIC\X_241_MS_ 430_5_3_unsigned.pdf · PDF p. 7.

Question 42 sources

2025 · Standard · 30/2/1 · Q23
Question paper: 041_Mathematics_Standard/30-2-1_Mathematics Standard.pdf · PDF p. 15.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-2-1_2024-25.pdf · PDF p. 8.

2025 · Standard · 30/2/2 · Q25
Question paper: 041_Mathematics_Standard/30-2-2_Mathematics Standard.pdf · PDF p. 15.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-2-2_2024-25.pdf · PDF p. 8.

2025 · Standard · 30/2/3 · Q24
Question paper: 041_Mathematics_Standard/30-2-3_Mathematics Standard.pdf · PDF p. 15.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-2-3_2024-25.pdf · PDF p. 8.

Question 43 sources

2026 · Standard · 30/5/1 · Q22
Question paper: Mathematics_Standard/1172-1_30-5-1 (Mathematics Standard).pdf · PDF p. 13.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-5-1_2025-26 (1).pdf · PDF p. 8.

2026 · Standard · 30/5/2 · Q25
Question paper: Mathematics_Standard/1172-2_30-5-2 (Mathematics Standard).pdf · PDF p. 13.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-5-2_2025-26 (1).pdf · PDF p. 9.

Question 44 sources

2026 · Basic · 430/8/1 · Q25
Question paper: Mathmatics_Basic/430-8-1_Maths Basic_E2_10_2026.pdf · PDF p. 15.
Marking scheme: Mathematics_Basic/241_430_8_1 Mathematics Basic.pdf · PDF p. 9.

2026 · Basic · 430/8/2 · Q22
Question paper: Mathmatics_Basic/430-8-2_Maths Basic_E2_10_2026.pdf · PDF p. 13.
Marking scheme: Mathematics_Basic/241_430_8_2 Mathematics Basic.pdf · PDF p. 9.

Question 45 sources

2024 · Basic · 430/B · Q23
Question paper: MATHEMATICS_BASIC/430(B) Maths Basic for VI.pdf · PDF p. 15.
Marking scheme: Mathematics/X_241_MS_430(B)_Unsigned.pdf · PDF p. 8.

Question 46 sources

2023 · Basic · 430/4/1 · Q21
Question paper: MATHEMATICS_BASIC/430_4_1 Maths Basic.pdf · PDF p. 5.
Marking scheme: Math_Basic/430_4_1.pdf · PDF p. 6.

2023 · Basic · 430/4/2 · Q23
Question paper: MATHEMATICS_BASIC/430_4_2 Maths Basic.pdf · PDF p. 5.
Marking scheme: Math_Basic/430_4_2.pdf · PDF p. 6.

Question 47 sources

2023 · Basic · 430/B · Q25
Question paper: MATHEMATICS_BASIC/430_B_Maths Basic for VI candidates.pdf · PDF p. 15.
Marking scheme: Math_Basic/430_VISUALLY IMPAIRED MATHEMATICS BASIC FINAL.pdf · PDF p. 9.

Question 48 sources

2024 · Basic · 430/S/1 · Q24
Question paper: 430-S-1 MATHEMATICS BASIC.pdf · PDF p. 13.
Marking scheme: Maths Basic/X_241_MS_430_S_1.pdf · PDF p. 8.

2024 · Basic · 430/S/2 · Q22
Question paper: 430-S-2 MATHEMATICS BASIC.pdf · PDF p. 13.
Marking scheme: Maths Basic/X_241_MS_430_S_2.pdf · PDF p. 8.

2024 · Basic · 430/S/3 · Q25
Question paper: 430-S-3 MATHEMATICS BASIC.pdf · PDF p. 13.
Marking scheme: Maths Basic/X_241_MS_430_S_3.pdf · PDF p. 8.

Question 49 sources

2024 · Standard · 30/B · Q23
Question paper: 30(B) Mathematics Standard.pdf · PDF p. 15.
Marking scheme: MATHS STANDARD/MS_041_30(B)-S Mathematics 2023-24.pdf · PDF p. 9.

Question 50 sources

2025 · Basic · 430/S/1 · Q25
Question paper: Maths_Basic/430-S-1Math Basic.pdf · PDF p. 13.
Marking scheme: MATHEMATICS_BASIC/MS_X_Mathematics(Basic)_241_430_S_1.pdf · PDF p. 10.

2025 · Basic · 430/S/2 · Q24
Question paper: Maths_Basic/430-S-2Math Basic.pdf · PDF p. 13.
Marking scheme: MATHEMATICS_BASIC/MS_X_Mathematics(Basic)_241_430_S_2.pdf · PDF p. 10.

2025 · Basic · 430/S/3 · Q22
Question paper: Maths_Basic/430-S-3Math Basic.pdf · PDF p. 13.
Marking scheme: MATHEMATICS_BASIC/MS_X_Mathematics(Basic)_241_430_S_3.pdf · PDF p. 10.

Question 51 sources

2026 · Standard · 30/B · Q24
Question paper: Mathematics_Standard/1164_30(B)_Mathematics Standard.pdf · PDF p. 15.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30(B)_2025-26 (1).pdf · PDF p. 8.

Question 52 sources

2022–23 · Basic · Basic · Q24
Question paper: MathsBasic-SQP.pdf · PDF p. 4.
Marking scheme: MathsBasic-MS.pdf · PDF p. 2.

Question 53 sources

2026–27 · Basic · Basic · Q22
Question paper: MathsBasic-SQP.pdf · PDF p. 4.
Marking scheme: MathsBasic-MS.pdf · PDF p. 4.

Question 54 sources

2026–27 · Standard · Standard · Q23
Question paper: MathsStandard-SQP.pdf · PDF p. 4.
Marking scheme: MathsStandard-MS.pdf · PDF p. 4.

Question 55 sources

2026 · Basic · 430/4/1 · Q21
Question paper: Mathematics_Standard/1158-1_430-4-1 Mathematics (Basic).pdf · PDF p. 13.
Marking scheme: 241_MATHEMATICS BASIC\X_241_MS_430_4_1_unsigned.pdf · PDF p. 7.

Question 56 sources

2026 · Basic · 430/4/2 · Q21
Question paper: Mathematics_Standard/1158-2_430-4-2 Mathematics (Basic).pdf · PDF p. 13.
Marking scheme: 241_MATHEMATICS BASIC\X_241_MS_430_4_2_unsigned.pdf · PDF p. 7.

Question 57 sources

2026 · Basic · 430/4/3 · Q21
Question paper: Mathematics_Standard/1158-3_430-4-3 Mathematics (Basic).pdf · PDF p. 13.
Marking scheme: 241_MATHEMATICS BASIC\X_241_MS_430_4_3_unsigned.pdf · PDF p. 7.

Question 58 sources

2026 · Standard · 30/5/2 · Q22
Question paper: Mathematics_Standard/1172-2_30-5-2 (Mathematics Standard).pdf · PDF p. 13.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-5-2_2025-26 (1).pdf · PDF p. 8.

Question 59 sources

2024 · Standard · 30/S/3 · Q23
Question paper: 30-S-3 MATHEMATICS STANDARD.pdf · PDF p. 13.
Marking scheme: MATHS STANDARD/MS 041_30-S-3 Mathematics 2023-24.pdf · PDF p. 8.

2024 · Standard · 30/S/1 · Q25
Question paper: 30-S-1 MATHEMATICS STANDARD.pdf · PDF p. 13.
Marking scheme: MATHS STANDARD/MS 041_30-S-1 Mathematics 2023-24.pdf · PDF p. 8.

Question 60 sources

2025 · Basic · 430/B · Q24
Question paper: Maths_Basic/430(B)-S Maths Basic Blind.pdf · PDF p. 13.
Marking scheme: MATHEMATICS_BASIC/430_B Final.pdf · PDF p. 9.

Question 61 sources

2025 · Basic · 430/1/1 · Q24
Question paper: 241_Mathematics_Basic/430-1-1_Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_1_1.pdf · PDF p. 9.

2025 · Basic · 430/1/2 · Q22
Question paper: 241_Mathematics_Basic/430-1-2_Mathematics Basic.pdf · PDF p. 13.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_1_2.pdf · PDF p. 9.

2025 · Basic · 430/1/3 · Q25
Question paper: 241_Mathematics_Basic/430-1-3_Mathematics Basic.pdf · PDF p. 13.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_1_3.pdf · PDF p. 9.

Question 62 sources

2025 · Basic · 430/3/1 · Q24
Question paper: 241_Mathematics_Basic/430-3-1_Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_3_1.pdf · PDF p. 9.

2025 · Basic · 430/3/2 · Q22
Question paper: 241_Mathematics_Basic/430-3-2_Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_3_2.pdf · PDF p. 9.

2025 · Basic · 430/3/3 · Q25
Question paper: 241_Mathematics_Basic/430-3-3_Mathematics Basic.pdf · PDF p. 13.
Marking scheme: Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_3_3.pdf · PDF p. 9.

Question 63 sources

2025 · Standard · 30/S/1 · Q25
Question paper: Maths_Standard/30-S-1 Maths Standard.pdf · PDF p. 17.
Marking scheme: MATHEMATICS_STANDARD/MS_X_041_Mathematics_30-S-1_2024-25.pdf · PDF p. 9.

2025 · Standard · 30/S/2 · Q21
Question paper: Maths_Standard/30-S-2 Maths Standard.pdf · PDF p. 13.
Marking scheme: MATHEMATICS_STANDARD/MS_X_041_Mathematics_30-S-2_2024-25.pdf · PDF p. 10.

2025 · Standard · 30/S/3 · Q23
Question paper: Maths_Standard/30-S-3 Maths Standard.pdf · PDF p. 15.
Marking scheme: MATHEMATICS_STANDARD/MS_X_041_Mathematics_30-S-3_2024-25.pdf · PDF p. 9.

Question 64 sources

2022–23 · Standard · Standard · Q25
Question paper: MathsStandard-SQP.pdf · PDF p. 5.
Marking scheme: MathsStandard-MS.pdf · PDF p. 3.

Question 65 sources

2023–24 · Standard · Standard · Q24
Question paper: MathsStandard-SQP.pdf · PDF p. 5.
Marking scheme: MathsStandard-MS.pdf · PDF p. 2.

Question 66 sources

2024–25 · Basic · Basic · Q24
Question paper: MathsBasic-SQP.pdf · PDF p. 4.
Marking scheme: MathsBasic-MS.pdf · PDF p. 2.

Question 67 sources

2025–26 · Standard · Standard · Q22
Question paper: MathsStandard-SQP.pdf · PDF p. 4.
Marking scheme: MathsStandard-MS.pdf · PDF p. 4.

Question 68 sources

2025 · Standard · 30/B · Q25
Question paper: Maths_Standard/30(B)-S Maths Standard Blind.pdf · PDF p. 15.
Marking scheme: MATHEMATICS_STANDARD/MS_X_041_Mathematics_30(B)-S_2024-25.pdf · PDF p. 8.

Question 69 sources

2025 · Standard · 30/4/1 · Q24
Question paper: 041_Mathematics_Standard/30-4-1_Mathematics Standard.pdf · PDF p. 9.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard _30-4-1_2024-25_.pdf · PDF p. 8.

2025 · Standard · 30/4/2 · Q21
Question paper: 041_Mathematics_Standard/30-4-2_Mathematics Standard.pdf · PDF p. 9.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-4-2_2024-25_.pdf · PDF p. 8.

2025 · Standard · 30/4/3 · Q23
Question paper: 041_Mathematics_Standard/30-4-3_Mathematics Standard.pdf · PDF p. 9.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-4-3_2024-25_.pdf · PDF p. 8.

Question 70 sources

2025 · Standard · 30/6/1 · Q22
Question paper: 041_Mathematics_Standard/30-6-1_Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-6-1_2024-25.pdf · PDF p. 8.

2025 · Standard · 30/6/2 · Q24
Question paper: 041_Mathematics_Standard/30-6-2_Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-6-2_2024-25.pdf · PDF p. 8.

2025 · Standard · 30/6/3 · Q21
Question paper: 041_Mathematics_Standard/30-6-3_Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-6-3_2024-25.pdf · PDF p. 8.

Question 71 sources

2023 · Standard · 30/C/B · Q25
Question paper: Maths_Standard/30_C_B Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Mathematics_Standard/041_30-C-B.pdf · PDF p. 9.

Question 72 sources

2023 · Standard · 30/5/1 · Q25
Question paper: MATHEMATICS_STANDARD/30_5_1_Maths Standard.pdf · PDF p. 13.
Marking scheme: Maths_Standard/MS 041_30-5-1 Mathematics 2022-23.pdf · PDF p. 11.

2023 · Standard · 30/5/2 · Q23
Question paper: MATHEMATICS_STANDARD/30_5_2_Maths Standard.pdf · PDF p. 11.
Marking scheme: Maths_Standard/MS 041_30-5-2 Mathematics 2022-23.pdf · PDF p. 12.

2023 · Standard · 30/5/3 · Q22
Question paper: MATHEMATICS_STANDARD/30_5_3_Maths Standard.pdf · PDF p. 11.
Marking scheme: Maths_Standard/MS 041_30-5-3 Mathematics 2022-23.pdf · PDF p. 11.

Question 73 sources

2023 · Basic · 430/C/B · Q24
Question paper: Maths_Basic/430_C_B Mathematics Basic.pdf · PDF p. 15.
Marking scheme: Mathematics_Basic/241_C_B.pdf · PDF p. 5.

Question 74 sources

2025 · Standard · 30/B · Q24
Question paper: 041_Mathematics_Standard/30 B_Maths Std for Visually Impaired.pdf · PDF p. 15.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30 (B)_2024-25.pdf · PDF p. 8.

Question 75 sources

2026 · Basic · 430/B · Q24
Question paper: Mathematics_Standard/1167_430(B)_Mathematics (Basic).pdf · PDF p. 13.
Marking scheme: 241_MATHEMATICS BASIC\X_241_MS_430(B)_unsigned.pdf · PDF p. 7.

Question 76 sources

2026 · Basic · 430/8/3 · Q24
Question paper: Mathmatics_Basic/430-8-3_Maths Basic_E2_10_2026.pdf · PDF p. 15.
Marking scheme: Mathematics_Basic/241_430_8_3 Mathematics Basic.pdf · PDF p. 9.

Question 77 sources

2023 · Standard · 30/1/3 · Q21
Question paper: MATHEMATICS_STANDARD/30_1_3_Maths Standard.pdf · PDF p. 13.
Marking scheme: Maths_Standard/MS 041_30-1-3 Mathematics 2022-23.pdf · PDF p. 7.

Question 78 sources

2024 · Standard · 30/B · Q24
Question paper: MATHEMATICS_STANDARD/30(B) Mathematics.pdf · PDF p. 13.
Marking scheme: Mathematics_Standard/MS 041_30(B) Mathematics 2023-24.pdf · PDF p. 9.

Question 79 sources

2025 · Standard · 30/3/3 · Q21
Question paper: 041_Mathematics_Standard/30-3-3_Mathematics Standard.pdf · PDF p. 13.
Marking scheme: Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-3-3_2024-25.pdf · PDF p. 8.

Question 80 sources

2026 · Basic · 430/B · Q23
Question paper: Mathmatics_Basic/430-B_Maths Basic_E2_10_2026.pdf · PDF p. 13.
Marking scheme: Mathematics_Basic/Mathematics Basic-241 E2 430(B).pdf · PDF p. 7.

Question 81 sources

2026 · Basic · 430/B · Q24
Question paper: Mathmatics_Basic/430-B_Maths Basic_E2_10_2026.pdf · PDF p. 13.
Marking scheme: Mathematics_Basic/Mathematics Basic-241 E2 430(B).pdf · PDF p. 7.

Question 82 sources

2026 · Standard · 30/4/1 · Q23
Question paper: Mathematics_Standard/1171-1_30-4-1 (Mathematics Standard).pdf · PDF p. 13.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-4-1_2025-26 (1).pdf · PDF p. 9.

2026 · Standard · 30/4/2 · Q25
Question paper: Mathematics_Standard/1171-2_30-4-2 (Mathematics Standard).pdf · PDF p. 13.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-4-2_2025-26 (1).pdf · PDF p. 9.

Question 83 sources

2026 · Standard · 30/4/3 · Q21
Question paper: Mathematics_Standard/1171-3_30-4-3 (Mathematics Standard).pdf · PDF p. 13.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-4-3_2025-26 (1).pdf · PDF p. 8.

Question 84 sources

2026 · Standard · 30/2/1 · Q24
Question paper: Mathematics_Standard/30_2_1_Mathematics Std._cl.10.pdf · PDF p. 17.
Marking scheme: 041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-2-1_2025-26 (1).pdf · PDF p. 9.

Build confidence in trigonometry

Get personal help with identities, angle equations and clear exam working.

Book a Free Assessment