Notes Formulae Read NCERT Chapter 8 (jemh108) Practice Check my answer NCERT Exemplar Solutions 15 NCERT textbook questions from Chapter 8
Exercise 8.1 In ∆ABC, right-angled at B, AB = 24 cm and BC = 7 cm. Determine sin A and cos A.
Stuck? Show a hint Step 1 Right-angled at B, so AC is the hypotenuse. By Pythagoras: AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625.
Step-by-step solution Right-angled at B, so AC is the hypotenuse. By Pythagoras: AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625. AC = 25 cm. For angle A: side opposite = BC = 7; side adjacent = AB = 24; hypotenuse = AC = 25. sin A = opposite / hypotenuse = BC / AC = 7/25. cos A = adjacent / hypotenuse = AB / AC = 24/25. Answer: sin A = 7/25, cos A = 24/25.
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If sin A = 3/4, calculate cos A and tan A.
Stuck? Show a hint Step 1 sin A = opposite/hypotenuse = 3/4, so let opposite = 3k and hypotenuse = 4k for some k > 0.
Step-by-step solution sin A = opposite/hypotenuse = 3/4, so let opposite = 3k and hypotenuse = 4k for some k > 0. By Pythagoras, adjacent² = hyp² − opp² = (4k)² − (3k)² = 16k² − 9k² = 7k². Adjacent = k√7. cos A = adjacent/hypotenuse = k√7 / 4k = √7 / 4. tan A = opposite/adjacent = 3k / (k√7) = 3/√7. Answer: cos A = √7/4, tan A = 3/√7.
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Given 15 cot A = 8, find sin A and sec A.
Stuck? Show a hint Step 1 15 cot A = 8 ⇒ cot A = 8/15, so tan A = 15/8.
Step-by-step solution 15 cot A = 8 ⇒ cot A = 8/15, so tan A = 15/8. tan A = opposite/adjacent = 15/8: let opposite = 15k, adjacent = 8k. Hypotenuse² = (15k)² + (8k)² = 225k² + 64k² = 289k² ⇒ hypotenuse = 17k. sin A = opposite/hypotenuse = 15k/17k = 15/17. sec A = hypotenuse/adjacent = 17k/8k = 17/8. Answer: sin A = 15/17, sec A = 17/8.
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Given sec θ = 13/12, calculate all other trigonometric ratios.
Stuck? Show a hint Step 1 sec θ = hypotenuse/adjacent = 13/12, so let hypotenuse = 13k and adjacent = 12k.
Step-by-step solution sec θ = hypotenuse/adjacent = 13/12, so let hypotenuse = 13k and adjacent = 12k. By Pythagoras, opposite² = 13²k² − 12²k² = (169 − 144)k² = 25k² ⇒ opposite = 5k. cos θ = 1/sec θ = 12/13. sin θ = opposite/hypotenuse = 5k/13k = 5/13. tan θ = opposite/adjacent = 5k/12k = 5/12; cot θ = 12/5. cosec θ = hypotenuse/opposite = 13k/5k = 13/5. Answer: sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, cosec θ = 13/5.
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NCERT Ex 8.1 Q7(i) 3 marks If cot θ = 7/8, evaluate (1 + sin θ)(1 − sin θ) / [(1 + cos θ)(1 − cos θ)].
Stuck? Show a hint Step 1 (1 + sin θ)(1 − sin θ) = 1 − sin²θ = cos²θ.
NCERT Ex 8.1 Q9(i) 2 marks In ∆ABC, right-angled at B, if tan A = 1/√3, find sin A cos C + cos A sin C.
Stuck? Show a hint Step 1 tan A = 1/√3 ⇒ ∠A = 30°, so ∠C = 60° (since the triangle is right-angled at B).
Which of the following statements is true?
(a) The value of tan A is always less than 1. (b) sec A = 12/5 for some value of angle A. (c) cos A is the abbreviation used for the cosecant of angle A. (d) sin θ = 4/3 for some angle θ. Stuck? Show a hint Step 1 (a) False — e.g. tan 60° = √3 > 1.
Exercise 8.2 NCERT Ex 8.2 Q1(i) 2 marks Evaluate sin 60° cos 30° + sin 30° cos 60°.
Stuck? Show a hint Step 1 sin 60° = √3/2, cos 30° = √3/2 ⇒ sin 60° · cos 30° = (√3/2)(√3/2) = 3/4.
NCERT Ex 8.2 Q1(ii) 3 marks Evaluate 2 tan²45° + cos²30° − sin²60°.
Stuck? Show a hint Step 1 tan 45° = 1 ⇒ 2 tan²45° = 2(1)² = 2.
The value of (2 tan 30°) / (1 + tan²30°) equals:
(a) sin 60° (b) cos 60° (c) tan 60° (d) sin 30° Stuck? Show a hint Step 1 tan 30° = 1/√3, so 2 tan 30° = 2/√3.
If tan(A + B) = √3 and tan(A − B) = 1/√3; 0° < A + B ≤ 90°, A > B, find A and B.
Stuck? Show a hint Step 1 tan(A + B) = √3 ⇒ A + B = 60°.
Exercise 8.3 9 sec²A − 9 tan²A equals:
(a) 1 (b) 9 (c) 8 (d) 0 Stuck? Show a hint Step 1 Factor: 9 sec²A − 9 tan²A = 9(sec²A − tan²A).
NCERT Ex 8.3 Q3(iii) 1 mark (sec A + tan A)(1 − sin A) equals:
(a) sec A (b) sin A (c) cosec A (d) cos A Stuck? Show a hint Step 1 (sec A + tan A)(1 − sin A) = (1/cos A + sin A/cos A)(1 − sin A) = [(1 + sin A)/cos A] · (1 − sin A).
NCERT Ex 8.3 Q4(ii) 5 marks Prove the identity: cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.
Stuck? Show a hint Step 1 Take LHS and combine over a common denominator cos A · (1 + sin A):
Step-by-step solution Take LHS and combine over a common denominator cos A · (1 + sin A): LHS = [cos²A + (1 + sin A)²] / [cos A (1 + sin A)]. Expand numerator: cos²A + 1 + 2 sin A + sin²A = (cos²A + sin²A) + 1 + 2 sin A = 1 + 1 + 2 sin A = 2 + 2 sin A. = 2(1 + sin A). So LHS = 2(1 + sin A) / [cos A (1 + sin A)] = 2/cos A = 2 sec A = RHS. Hence proved. Answer: LHS = 2 sec A = RHS.
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NCERT Ex 8.3 Q4(vi) 5 marks Prove the identity: √[(1 + sin A)/(1 − sin A)] = sec A + tan A.
Stuck? Show a hint Step 1 Rationalise inside the square root by multiplying numerator and denominator by (1 + sin A):
Step-by-step solution Rationalise inside the square root by multiplying numerator and denominator by (1 + sin A): (1 + sin A)/(1 − sin A) = (1 + sin A)² / [(1 − sin A)(1 + sin A)] = (1 + sin A)² / (1 − sin²A) = (1 + sin A)² / cos²A. Take the (positive) square root: √[(1 + sin A)² / cos²A] = (1 + sin A) / cos A (taking +ve root since A acute). = 1/cos A + sin A/cos A = sec A + tan A. Hence LHS = sec A + tan A = RHS. Proved. Answer: LHS = sec A + tan A = RHS.
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