Tiêu đề TRIGONOMETRIC FUNCTIONS A-3.pdf ✅

Class : XIth Subject : MATHS Date : DPP No. :3 741 (a) It is given that A,B,C are in A.P. ∴ 2B = A + C⇒3B = A + B + C⇒3B = 180°⇒B = 60° ⇒ cosB = 1 2 ⇒ c 2 + a 2 ― b 2 2ac = 1 2 ⇒c 2 + a 2 ― b 2 = ac ⇒(a ― c) 2 = b 2 ― ac ⇒|a ― c| = b 2 ― ac ⇒|sinA ― sin C| = sin2B ― sinA sin C ⇒2|sin A ― C 2 | cos A + C 2 = 3 4 ― sinA sin C ⇒2|sin A ― C 2 | = 3 ― 4 sinA sin C ⇒ 3 ― 4 sinA sin C |A ― C| = 2|sinA ― C 2 | |A ― C| ⇒ lim A→C 3 ― 4 sinA sin C |A ― C| = lim A→C | sin ( A ― C 2 ) A ― C 2 | = 1 742 (c) 3 ― cos θ + cos (θ + π 3 ) = 3 ― cos θ + 1 2 cos θ ― 3 2 sin θ = 3 ― 1 2 cos θ ― 3 2 sin θ = 3 ― sin (θ + π 6 ) Topic :-TRIGONOMETRIC FUNCTIONS Solutions
Since, ―1 ≤ sin θ ≤ 1 Hence, the value of expression lies in [2, 4] 743 (c) We have, cosA = m cosB ⇒ cosA cosB = m 1 ⇒ cosA + cosB cosA ― cosB = m + 1 m ― 1 ⇒ 2 cos A + B 2 cos B ― A 2 2 sinA + B 2 sinB ― A 2 = m + 1 m ― 1 ⇒ cot A + B 2 = ( m + 1 m ― 1 )tan B ― A 2 But cot A + B 2 = λtan B ― A 2 ∴ λ = m + 1 m ― 1 744 (c) cos4 π 8 + cos4 7π 8 + cos4 3π 8 + cos4 5π 8 = cos4 π 8 + cos4 π 8 + cos4 ( π 2 ― π 8 ) + cos4 ( π 2 + π 8 ) = 2[cos4 π 8 + sin4 π 8 ] = 2[(cos2 π 8 + sin2 π 8 ) 2 ― 2 sin2 π 8 cos2 π 8 ] = 2[1 ― 1 2 (sin π 4 ) 2 ] = 2[1 ― 1 4 ] = 3 2 745 (b) Given, sin θ = 12 13 and cos φ = ― 3 5 ∴ cos θ = 5 13 and sin φ = ― 4 5 ∴ sin(θ + φ) = sin θ cos φ + cos θ sin φ = 12 13 × ( ― 3 5 ) + 5 13 × ( ― 4 5 )
= ―36 65 + ( ― 20) 65 = ― 56 65 746 (c) We have, sec2 θ cosec 2θ = sin2 θ + cos2 θ sin2 θ cos2 θ ― 4 sin2 2θ ≥ 4 and, sec2 θcosec 2θ = 4 sin2 2θ ≥ 4 Thus, the required equation is x 2 ―λx + λ = 0, where λ ≥ 4 747 (a) 1 m[(m + n) + 1 (m + n)] = 1 sec θ [sec θ + tan θ + 1 sec θ + tan θ ] = [ sec2 θ + tan2 θ + 2 sec θ tan θ + 1 ] sec θ( sec θ + tan θ ) = 2 sec2 θ + 2 sec θ tan θ sec θ( sec θ + tan θ ) = 2 748 (a) ∵ sinA sinB = 1 2 × 2 sinA sinB = 1 2 [cos( A ― B) ― cos(A + B)] = 1 2 [cos( A ― B) ― cos 90°] ( ∵ A + B + C = 180° and ∠C = 90°, given) = 1 2 cos(A ― B) ≤ 1 2 ∴ Maximum value of sinA sinB = 1 2 749 (b) In cyclic quadrilateral ABCD, we have A + C = π and B + D = π ∴ cosA = ― cos C and cosB = ― cosD ⇒ cosA + cosB + cos C + cosD = 0 750 (d) Let A = θ,B = 2 θ and C = 3 θ. Then, A + B + C = 180°⇒6 θ = 180°⇒θ = 30° ∴ A = 30°,B = 60° and C = 90° Now,
a :b :c = sinA : sinB : sin C, ⇒a :b :c = 1 2 : 3 2 :1⇒a :b :c = 1 : 3 :2 751 (b) We have, sin(π cos θ ) = cos(π sin θ) ⇒ sin(π cos θ ) = sin ( π 2 + π sin θ) ⇒π cos θ = π 2 + π sin θ ⇒π cos θ ― π sin θ = π 2 ⇒ cos θ ― sin θ = 1 2 ⇒ 1 2 cos θ ― 1 2 sin θ = 1 2 2 ⇒ cos (θ + π 4 ) = 1 2 2 752 (a) The given equation is not meaningful for |cos x| = 1 So, let |cos x| ≠ 1 Now, 2 1+|cos x|+cos2 x+|cos x| 3+...+to ∞ = 4 ⇒ 1 2 1―| cos x | = 2 2 ⇒ 1 1 ― | cos x | = 2 ⇒2 ― 2|cos x| = 1 ⇒|cos x| = 1 2 ⇒ cos x =± 1 2 ⇒ cos x = cos π 3 , cos x = cos 2 π 3 ⇒x = 2n π ± π 3 ,x = 2n π ± 2 π 3 ,n ∈ Z ⇒x = 2n π ± π 3 ,x = (2n ± 1) π ± π 3 ⇒x = nπ ± π 3 ,n ∈ Z 753 (b) We have, (cosα + cosβ) 2 ― (sinα + sinβ) 2 = 0 ⇒(cos2α + cos2β + 2 cosα cosβ) ― (sin2α + sin2β + 2 sinα sinβ) = 0 ⇒ cos 2α + cos 2β = ―2(cosα cosβ ― sinα sinβ) ⇒ cos 2α + cos 2β = ―2 cos(α + β)