Tiêu đề TRIGONOMETRIC FUNCTIONS A-8.pdf ✅

Class : XIth Subject : MATHS Date : DPP No. :8 84 2 (a) Let ABC be the triangle such that a = 2 2 cm, b = 2 3 cm and ∠A = π 4 ∴ cosA = b 2 + c 2 ― a 2 2bc ⇒ 1 2 = 12 + c 2 ― 8 4 3 c ⇒4 + c 2 = 2 6 c ⇒c 2 ― 2 6 c + 4 = 0 ⇒c = 2 6 ± 24 ― 16 2 ⇒c = 6 ± 2⇒c = 6 + 2 [ ∵ c is the largest side] 84 3 (b) We have, rsin θ = 3 and r = 4(1 + sin θ) ⇒r = 4 + 12 r ⇒r 2 ― 4r ― 12 = 0 ⇒(r ― 6)(r + 2) = 0 ⇒r = 6 [ ∵ r > 0] ∴ r sin θ = 3⇒ sin θ = 1 2 ⇒θ = π 6 , 5π 6 Hence, the total number of ordered pairs of the form (r, θ) is 1 × 2 = 2 84 4 (d) We have, sin 65° + sin 43° ― sin 29° ― sin 7° = (sin 65° + sin 43°) ― (sin 29° + sin 7°) = 2 sin 54° cos 11° ― 2 sin 18° cos 11° = 2 cos 11°(cos 36° ― sin 18°) = 2 cos 11°( 5 + 1 4 ― 5 ― 1 4 ) = cos 11° 84 5 (c) We have, sinB = 1 5 sin(2A + B) ⇒ sin(2A + B) sinB = 5 1 ⇒ sin(2A + B) + sinB sin(2A + B) ― sinB = 5 + 1 5 ― 1 ⇒ 2 sin(A + B) cosA 2 sinA cos(A + B) = 3 2 ⇒ tan(A + B) tanA = 3 2 84 6 (b) Since, A + B + C = π ⇒a = π ― (B + C) We have, cosA = cosB cos C ⇒ cos[π ― (B + C)] = cosB cos C ⇒ ― cos( B + C) = cosB cos C ⇒ ― [cosB cos C ― sinB sin C] = cosB cos C ⇒ sinB sin C = 2 cosB cos C ⇒ tanB tan C = 2 84 7 (a) We have, sin x + cosec x = 2⇒(sin x ― 1) 2 = 0⇒ sin x = 1 ∴ sinn x + cosec n x = 1 + 1 = 2 84 8 (a) We have, a 2 ― b 2 a 2 + b 2 = sin(A ― B) sin(A + B) ⇒ sin2A ― sin2B sin2A + sin2B = sin2A ― sin2B sin2 (A + B) ⇒(sin2A ― sin2B)(sin2A + sin2B ― sin2 C) = 0 ⇒ sin(A + B) sin(A ― B)( sin2A + sin2B ― sin2 C ) = 0 ⇒sin(A ― B) = 0 or, sin2A + sin2B = sin2 C ⇒A = B or, a 2 + b 2 = c 2 ⇒∆ABC is either right angled or isosceles 84 9 (c) We have, cosA = cosB cos C ⇒ cos{π ― (B + C)} = cosB cos C ⇒ ― cos(B + C) = cosB cos C ⇒2 cosB cos C = sinB sin C ⇒ cotB cot C = 1 2 Topic :-TRIGONOMETRIC FUNCTIONS Solutions
85 0 (c) We have, 2s = a + b + c = 13 + 14 + 15 ⇒s = 21 ⇒s ― a = 8,s ― b = 7 and s ― c = 6 Now, 1 r1 : 1 r2 : 1 r3 = s ― a ∆ : s ― b ∆ : s ― c ∆ ⇒ 1 r1 : 1 r2 : 1 r3 = s ― a :s ― b :s ― c = 8 :7 :6 851 (a) We have, sin 7θ + 6 sin 5θ + 17 sin 3θ + 12 sin θ sin 6θ + 5 sin 4θ + 12 sin 2θ = (sin 7θ + sin 5θ) + 5(sin 5θ + sin 3θ) + 12(sin 3θ + sin θ) sin 6θ + 5 sin 4θ + 12 sin 2θ = 2 sin 6θ cos θ + 10 sin 4θ cos θ + 24 sin 2θ cos θ sin 6θ + 5 sin 4θ + 12 sin 2θ = 2 cos θ(sin 6θ + 5 sin 4θ + 12 sin 2θ) sin 6θ + 5 sin 4θ + 12 sin 2θ = 2 cos θ
852 (a) Let the angles be A = x ― d,B = x,C = x + d. Then, x ― d + x + x + d = 180°⇒x = 60° Therefore, two larger angles are B = 60° and C Hence, b = 9 and c = 10 Now, cosB = c 2 + a 2 ― b 2 2 ac ⇒ 1 2 = 100 + a 2 ― 81 20 a ⇒a 2 ― 10 a + 19 = 0⇒a = 5 ± 6 853 (b) Since, cos 2x, 1 2 , sin 2x are in AP ⇒ cos 2x + sin 2x = 1 ⇒ sin 2x = 1 ― cos 2x = 2 sin2 x ⇒ 2 sin x( cos x ― sin x) = 0 ⇒ sin x = 0 or cos x ― sin x = 0 ⇒ x = nπ or tan x = 1 ⇒ x = nπ or x = nπ + π 4 Thus, required values of x are nπ and nπ + π 4 854 (b) cos π 18 + cos 2π 18 + ... + cos 16π 18 + cos 17π 18 + cos π = cos π 18 + cos 2π 18 + ... ― cos 2π 18 ― cos π 18 + cos π = cos π = ―1 855 (b) Given that, sin θ + cos θ = 1 ⇒ 1 2 sin θ + 1 2 cos θ = 1 2 ⇒ sin ( θ + π 4 ) = 1 2 = sin π 4 ⇒ θ + π 4 = nπ + ( ―1) n π 4 ⇒ θ = nπ + ( ―1) n π 4 ― π 4 856 (d) We have, 0 < x < π⇒ sin x > 0 Now, 1 + sin x + sin2 x + ...∞ = 4 + 2 3 ⇒ 1 1 ― sin x = 4 + 2 3
⇒ sin x = 1 ― 1 4 + 2 3 ⇒ sin x = 3 + 2 3 4 + 2 3 = 3 2 ⇒x = π 3 or, 2π 3 857 (b) sin x + sin y cos x + cos y = a b ⇒ 2 sin ( x + y 2 ) cos ( x ― y 2 ) 2 cos ( x + y 2 ) cos ( x ― y 2 ) = a b ⇒ tan ( x + y 2 ) = a b 858 (b) The given equation can be written as cos(π tan θ ) = cos ( π 2 ― π cot θ) ⇒π tan θ = 2n π ± ( π 2 ― π cot θ) ,n ∈ Z ⇒ tan θ = 2n ± ( 1 2 ― cot θ), n ∈ Z ⇒ tan θ ― cot θ = 2n ― 1 2 ,n ∈ Z [Taking negative sign] ⇒ tan2 θ ― 1 tan θ = 2n ― 1 2 ⇒ tan2 θ ― 1 2 tan θ = n ― 1 4 ⇒ 1 ― tan2 θ 2 tan θ = ―n + 1 4 ⇒ cot 2 θ = m + 1 4 , where m = ―n ∈ Z 859 (c) From Questions 47, we have ∆ = 1 2 ap1,∆ = 1 2 bp2,∆ = 1 2 cp3 Now, p1,p2,p3 are in A.P. ⇒ 2 ∆ a , 2 ∆ b , 2 ∆ c are in A.P. ⇒ 1 a , 1 b , 1 c are in A.P.⇒a,b,c are in H.P. 860 (d)