Tiêu đề TRIGONOMETRIC FUNCTIONS A-4.pdf ✅

Class : XIth Subject : MATHS Date : DPP No. :4 761 (c) We have, a = sin4 θ + cos4 θ ≤ sin2 θ + cos2 θ ≤ 1 Also, a = sin4 θ + cos4 θ = (sin2 θ + cos2 θ) 2 ― 2 sin2 θ + cos2 θ ⇒a = sin4 θ + cos4 θ = 1 ― 1 2 sin2 2θ ⇒ sin2 2θ = 2(1 ― a)⇒2(1 ― a) ≤ 1⇒a ≥ 1 2 Hence, 1 2 ≤ a ≤ 1 762 (a) Let 3 +1 = rcosα and 3 ― 1 = r sin α, then r = ( 3 + 1) 2 + ( 3 ― 1) 2 = 3 + 1 + 2 3 + 3 + 1 ― 2 3 = 2 2 and tan α = 3 ― 1 3 + 1 = 1 ― ( 1 3 ) 1 + ( 1 3 ) = tan ( π 4 ― π 6 ) ⇒ α = π 12 The given equation reduces to 2 2cos(θ ― α) = 2 ⇒ cos ( θ ― π 12) = cos π 4 ⇒ θ ― π 12 = 2nπ ± π 4 ⇒ θ = 2nπ ± π 4 + π 12 763 (d) sin(A + B) = sinA cosB + sinB cosA = 1 10 . 1 ― 1 5 + 1 5 1 ― 1 10 Topic :-TRIGONOMETRIC FUNCTIONS Solutions
[ ∵ sinA = 1 10 , sinB = 1 5 ] = 1 10 4 5 + 1 5 9 10 = 5 50 = 1 2 = sin π 4 ⇒ A + B = π 4 764 (b) The given equation can be rewritten as tan θ( sin θ + 3) = 0 ⇒tan θ = 0, but sin θ + 3 ≠ 0 ⇒ tan θ = 0 ⇒ θ = nπ, n ∈ I 765 (a) We have, y = sin θ ― cos θ and sin θ ― cos θ lies between ― 2 and + 2 ∴ ― 2 ≤ y ≤ 2 766 (c) Now,sin(α ― β) = sin(θ ― β ― (θ ― α)) = sin(θ ― β) = cos(θ ― α) ― cos(θ ― β)sin(θ ― α) = ba ― 1 ― b 2 1 ― a 2 and cos (α ― β) = cos(θ ― β ― (θ ― α)) = cos(θ ― β)cos(θ ― α) + sin(θ ― β)sin(θ ― α) = a 1 ― b 2 + b 1 ― a 2 ∴ cos2 (α ― β) + 2ab sin(α ― β) = (a 1 ― b 2 + b 1 ― a 2 ) 2 + 2ab(ab ― 1 ― a 2 1 ― b 2 ) = a 2 + b 2 767 (d) We have, 8 sec2 θ ― 6 sec θ + 1 = 0 ⇒(4 sec θ ― 1)(2 sec θ ― 1) = 0 ⇒ sec θ = 1 4 , sec θ = 1 2 But, this is not possible as |sec θ| ≥ 1 768 (c) We have, x 3 ― 13x 2 + 54x ― 72 = 0 ⇒(x ― 3)(x 2 ― 10x + 24) = 0
⇒(x ― 3)(x ― 4)(x ― 6) = 0⇒x = 3,4,6 Let a = 3,b = 4 and c = 6 ∴ cosA a + cosB b + cos C c = a 2 + b 2 + c 2 2abc = 61 144 769 (b) cos4 θ ― sin4 θ = (cos2 θ ― sin2 θ)( cos2 θ + sin2 θ ) = cos 2θ = 2 cos2 θ ― 1 770 (b) cos 15° cos 7 1 2 ° sin 7 1 2 ° = 1 2 cos 15° sin 15° = 1 4 sin 30° = 1 4 × 1 2 = 1 8 771 (b) 1 ― sin θ 1 + sin θ + 1 + sin θ 1 ― sin θ = 1 ― sin θ + 1 + sin θ 1 ― sin2 θ = 2 cos2 θ = 2 | cos θ | = ― 2 cos θ = ―2 sec θ( ∵ π 2 < θ < π) 772 (c) cos2A(3 ― 4 cos2A) 2 + sin2A(3 ― 4 sin2A) 2 = (3 cosA ― 4 cos3A) 2 + (3 sinA ― 4 sin3A) 2 = ( ― cos 3A) 2 + (sin 3A) 2 = 1 773 (a) It is given that A,B,C are in A.P. ∴ 2B = A + C ⇒3B = A + B + C⇒3B = 180°⇒B = 60° Also, b :c = 3 : 2 ⇒ sinB sin C = 3 2 ⇒ 3 2 sin C = 3 2 ⇒ sin C = 1 2 ⇒C = 45° ∴ A = 180° ― (60° + 45°) = 75° 774 (b) We have, tan x = b a
∴ a + b a ― b + a ― b a + b = 1 + b a 1 ― b a + 1 ― b a 1 + b a = 1 + tan x 1 ― tan x + 1 ― tan x 1 + tan x = cos x + sin x cos x ― sin x + cos x ― sin x cos x + sin x = cos x + sin x + cos x ― sin x cos2 x ― sin2 x = 2 cos x cos 2x 775 (c) We have, sinA + cosA = m and sin3A + cos3A = n Now, sinA + cosA = m ⇒(sinA + cosA) 3 = m3 ⇒ sin3A + cos3A + 3 sinA cosA(sinA + cosA) = m3 ⇒n + 3sinAcosA m = m3 ...(i) Again, sinA + cosA = m ⇒ sin2A + cos2A + 2 sinA cosA = m2 ⇒sinAcosA = m2 ― 1 2 ...(ii) From (i) and (ii), we have n + 3m (m2 ― 1) 2 = m3 ⇒2n + 3m3 ― 3m = 2m3⇒m3 ― 3m + 2n = 0 776 (b) ∵ sec x ― 1 = ( 2 ― 1)tan x ⇒ 1 ― cos x = ( 2 ― 1)sin x ⇒ sin x 2 {sin x 2 ― ( 2 ― 1)cos x 2 } = 0 ⇒ sin x 2 = 0 or tan x 2 = 2 ― 1 = tan π 8 ⇒ x 2 = nπ or x 2 = nπ + π 8 ∴ x = 2nπ, 2nπ + π 4 777 (a) α ― β = (θ ― β) ― (θ ― α) ∴ cos(α ― β) = cos(θ ― β)cos(θ ― α) + sin(θ ― β) sin(θ ― α)