Chapter 8, Introduction to Trigonometry, is one of the highest-weightage chapters in the Class 10 Maths board exam — it introduces the six trigonometric ratios, their values for standard angles, and the identities that Chapter 9 (Applications of Trigonometry) builds on directly.
This post covers the current 2026-27 rationalised exercise set: Exercise 8.1 (11 questions), Exercise 8.2 (4 questions) and Exercise 8.3 (complementary-angle identities). Every answer below is derived step by step from the ratio definitions, not copied from any answer key.
Exercise 8.1
Q1. In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine (i) sin A, cos A (ii) sin C, cos C.
Solution: By Pythagoras, AC² = 24² + 7² = 625, so AC = 25 cm.
sin A = 7/25, cos A = 24/25. sin C = 24/25, cos C = 7/25.
Q2. In ΔPQR, right-angled at Q, PR = 13 cm, PQ = 12 cm. Find tan P − cot R.
Solution: QR² = 169 − 144 = 25, QR = 5 cm. tan P = 5/12, cot R = 5/12. tan P − cot R = 0.
Q3. If sin A = 3/4, calculate cos A and tan A.
Solution: adjacent = √(16−9) = √7. cos A = √7/4, tan A = 3/√7.
Q4. Given 15 cot A = 8, find sin A and sec A.
Solution: cot A = 8/15, hypotenuse = √(64+225) = 17. sin A = 15/17, sec A = 17/8.
Q5. Given sec θ = 13/12, calculate all other trigonometric ratios.
Solution: opposite = √(169−144) = 5. sin θ=5/13, cos θ=12/13, tan θ=5/12, cosec θ=13/5, cot θ=12/5.
Q6. If ∠A and ∠B are acute angles with cos A = cos B, show ∠A = ∠B.
Solution: Cosine is one-to-one and strictly decreasing on 0°–90°, so equal cosine values force equal angles.
Q7. If cot θ = 7/8, evaluate (i) [(1+sinθ)(1−sinθ)]/[(1+cosθ)(1−cosθ)] (ii) cot²θ.
Solution: (1−sin²θ)/(1−cos²θ) = cos²θ/sin²θ = cot²θ = 49/64 for both parts.
Q8. If 3 cot A = 4, check (1−tan²A)/(1+tan²A) = cos²A−sin²A.
Solution: tan A = 3/4, sin A=3/5, cos A=4/5. LHS = (7/16)/(25/16)=7/25. RHS = 16/25−9/25=7/25. Verified equal.
Q9. In ΔABC, right-angled at B, tan A = 1/√3. Find (i) sinAcosC+cosAsinC (ii) cosAcosC−sinAsinC.
Solution: A=30°, C=60°. (i) sin(A+C)=sin90°=1. (ii) cos(A+C)=cos90°=0.
Q10. In ΔPQR, right-angled at Q, PR+QR=25 cm, PQ=5 cm. Find sinP, cosP, tanP.
Solution: (PR−QR)(PR+QR)=PQ²=25 ⇒ PR−QR=1, so PR=13, QR=12. sinP=12/13, cosP=5/13, tanP=12/5.
Q11. True/False: (i) tan A always <1 — False. (ii) sec A=12/5 possible — True. (iii) cos A abbreviates cosecant — False. (iv) cot A is a product of cot and A — False. (v) sinθ=4/3 possible — False (sinθ ≤ 1 always).
Exercise 8.2
Q1. Evaluate: (i) sin60°cos30°+sin30°cos60° = 1. (ii) 2tan²45°+cos²30°−sin²60° = 2. (iii) cos45°/(sec30°+cosec30°) = (3√2−√6)/8. (iv) (sin30°+tan45°−cosec60°)/(sec30°+cos60°+cot45°) = (43−24√3)/11. (v) (5cos²60°+4sec²30°−tan²45°)/(sin²30°+cos²30°) = 67/12.
Q2. Choose the correct option: (i) (2tan30°)/(1+tan²30°) = sin60°. (ii) (1−tan²45°)/(1+tan²45°) = 0. (iii) sin2A=2sinA true when A=0°. (iv) (2tan30°)/(1−tan²30°) = tan60°.
Q3. If tan(A+B)=√3 and tan(A−B)=1/√3, find A and B.
Solution: A+B=60°, A−B=30°. A=45°, B=15°.
Q4. True/False: (i) sin(A+B)=sinA+sinB — False. (ii) sinθ increases as θ increases — True. (iii) cosθ increases as θ increases — False. (iv) sinθ=cosθ for all θ — False. (v) cot A undefined at A=0° — True.
Exercise 8.3 — complementary angles
Q1. Evaluate (i) sin18°/cos72° (ii) tan26°/cot64° (iii) cos48°−sin42° (iv) cosec31°−sec59°.
Solution: Each pair is complementary, so each expression collapses to a ratio over itself or a difference of equals: answers are 1, 1, 0, 0.
Q2. Show (i) tan48°tan23°tan42°tan67°=1 (ii) cos38°cos52°−sin38°sin52°=0.
Solution: (i) tan48°·tan42°=1 and tan23°·tan67°=1 by complementary pairing, product=1. (ii) equals cos(38°+52°)=cos90°=0.
Q3. tan2A=cot(A−18°), find A. Solution: 90°−2A=A−18° ⇒ A=36°.
Q4. If tanA=cotB, prove A+B=90°. Solution: cotB=tan(90°−B) ⇒ A=90°−B.
Q5. sec4A=cosec(A−20°), find A. Solution: 90°−4A=A−20° ⇒ A=22°.
Q6. Show sin((B+C)/2)=cos(A/2) for ΔABC. Solution: B+C=180°−A ⇒ (B+C)/2=90°−A/2.
Q7. Express sin67°+cos75° in terms of angles below 45°. Solution: = cos23°+sin15°.
Why this chapter matters for boards
These ratios and complementary-angle identities reappear directly in Chapter 9’s heights-and-distances problems and show up as standalone 2-mark questions across the paper — spotting A+(90°−A) pairs quickly is worth practising until automatic.
FAQ
Is Class 10 Trigonometry hard? Not if the ratio definitions and standard-angle values (0°,30°,45°,60°,90°) are memorised first.
How many exercises does Chapter 8 have for 2026-27? Three — 8.1 (11 questions), 8.2 (4 questions), 8.3 (complementary-angle identities).
See every chapter’s official PDF links on the https://ncertbooks.org/ncert-books/class-10-mathematics-ncert-book-pdf/ page. For harder practice see the https://ncertbooks.org/ncert-extra-questions-class-10-maths-chapter-8-trigonometry/ post, for quick revision see https://ncertbooks.org/ncert-revision-notes-class-10-maths-chapter-8-trigonometry/, and every formula from this chapter is indexed in the https://ncertbooks.org/ncert-class-10-maths-formulas-handbook/.
Part of the Class 10 Maths Chapter 8 cluster: Solutions | Extra Questions (HOTS) | Revision Notes | Formulas Handbook | Class 10 Maths book page.