Exercise 8.1 11 Questions – Finding Trigonometric Ratios
In Î"ABC right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
By Pythagoras: AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 â‡' AC = 25 cm
For ∠A: Opp=BC=7, Adj=AB=24, Hyp=25. sin A=7/25, cos A=24/25
For ∠C: Opp=AB=24, Adj=BC=7, Hyp=25. sin C=24/25, cos C=7/25
Q2: If sin A = 3/4, find cos A and tan A. â†' cos A = √(1âˆ'9/16) = √7/4. tan A = 3/√7
Q3: If cot θ = 7/8, find (1+sinθ)(1âˆ'sinθ)/(1+cosθ)(1âˆ'cosθ). â†' sinθ=8/√113, cosθ=7/√113. Answer = 49/64
Q5: Given sec θ = 13/12, find all other trigonometric ratios. â†' cos θ=12/13, sin θ=5/13, tan θ=5/12, cot θ=12/5, cosec θ=13/5
Q6: If ∠A and ∠B are acute such that cos A = cos B, prove ∠A = ∠B. â†' Since cosine is decreasing in (0,90°), equal cosine â‡' equal angle.
Q8: If 3 cot A = 4, check whether (1âˆ'tan²A)/(1+tan²A) = cos²A âˆ' sin²A. â†' tan A = 3/4, sin A = 3/5, cos A = 4/5. LHS = (1âˆ'9/16)/(1+9/16) = 7/25. RHS = 16/25âˆ'9/25 = 7/25 âœ"
Q10: In Î"PQR right at Q, PR+QR=25cm, PQ=5cm. Find sin P, cos P, tan P. â†' PR²=PQ²+QR². PR=QR+25âˆ'PR. Solve: PR=13, QR=12. sin P=12/13, cos P=5/13, tan P=12/5
Exercise 8.2 4 Questions – Standard Angle Values
Evaluate: (i) sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4+1/4 = 1
(ii) 2 tan² 45° + cos² 30° âˆ' sin² 60° = 2(1)+(3/4)âˆ'(3/4) = 2
Q2: Choose correct option: (i) 2 tan 30°/(1+tan²30°) = 2(1/√3)/(1+1/3) = (2/√3)×(3/4) = √3/2 = sin 60°
Q4: State True/False: (i) sin(A+B)=sinA+sinB â†' False. (ii) sinθ increases as θ increases from 0° to 90° â†' True. (iii) cosθ decreases as θ increases â†' True. (iv) sinθ=cosθ for all θ â†' False (only at 45°). (v) cot A is not defined for A=0° â†' True.
Exercise 8.3 7 Questions – Trigonometric Identities
(i) (cosec θ âˆ' cot θ)² = (1 âˆ' cos θ)/(1 + cos θ)
LHS = (1/sinθ âˆ' cosθ/sinθ)² = ((1âˆ'cosθ)/sinθ)² = (1âˆ'cosθ)²/sin²θ = (1âˆ'cosθ)²/(1âˆ'cos²θ) = (1âˆ'cosθ)/(1+cosθ) = RHS âœ"
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
LHS = [cos²A + (1+sinA)²] / [cosA(1+sinA)] = [cos²A+1+2sinA+sin²A] / [cosA(1+sinA)] = [2+2sinA] / [cosA(1+sinA)] = 2(1+sinA)/[cosA(1+sinA)] = 2/cosA = 2secA = RHS âœ"
Q4: Prove (1+secA)/secA = sin²A/(1âˆ'cosA). Q5: (sinA+cosecA)²+(cosA+secA)² = 7+tan²A+cot²A
Q7: Express sin 67°+cos 75° in terms of ratios of angles between 0° and 45°. â†' sin 67° = cos 23°, cos 75° = sin 15°. So cos 23° + sin 15°
Exercise 8.4 5 Questions – Advanced Identities
Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A. â†' sin A = 1/√(1+cot²A), sec A = √(1+cot²A)/cot A, tan A = 1/cot A
Q2: Write all other ratios of ∠A in terms of sec A.
Q4: Prove (1+tan²A)/(1+cot²A) = (1âˆ'tanA/1âˆ'cotA)² = tan²A. Choose correct option â†' (D) tan²A
Q5: Prove: √(1+sinA)/(1âˆ'sinA) = secA + tanA. LHS = √(1+sinA)²/(1âˆ'sin²A) = (1+sinA)/cosA = secA + tanA = RHS âœ"
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