Question

40 cosec theta is equal to 41,then cos theta​

Answer 1

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Given: 9 sin θ + 40 cos θ= 41

Given: 9 sin θ + 40 cos θ= 41 ⇒ 9 sin θ = 41 – 40 cosθ

Given: 9 sin θ + 40 cos θ= 41 ⇒ 9 sin θ = 41 – 40 cosθ …(i) Squaring both sides, we get

Given: 9 sin θ + 40 cos θ= 41 ⇒ 9 sin θ = 41 – 40 cosθ …(i) Squaring both sides, we get ⇒ 81sin2 θ = 1681+1600 cos2 θ – 2(41) (40cos θ) [∵ (a – b)2 =a2 +b2 –2ab]

Given: 9 sin θ + 40 cos θ= 41 ⇒ 9 sin θ = 41 – 40 cosθ …(i) Squaring both sides, we get ⇒ 81sin2 θ = 1681+1600 cos2 θ – 2(41) (40cos θ) [∵ (a – b)2 =a2 +b2 –2ab] ⇒ 81 (1– cos2 θ) =1681+1600 cos2 θ – 3280cosθ ⇒ 81 – 81cos2 θ = 1681 +1600cos2 θ – 3280 cosθ ⇒ 1681cos2 θ –3280cos θ +1600 = 0

Given: 9 sin θ + 40 cos θ= 41 ⇒ 9 sin θ = 41 – 40 cosθ …(i) Squaring both sides, we get ⇒ 81sin2 θ = 1681+1600 cos2 θ – 2(41) (40cos θ) [∵ (a – b)2 =a2 +b2 –2ab] ⇒ 81 (1– cos2 θ) =1681+1600 cos2 θ – 3280cosθ ⇒ 81 – 81cos2 θ = 1681 +1600cos2 θ – 3280 cosθ ⇒ 1681cos2 θ –3280cos θ +1600 = 0 ⇒ (41)2 cos2 θ – 2(41) (40cos θ) + (40)2 = 0

Given: 9 sin θ + 40 cos θ= 41 ⇒ 9 sin θ = 41 – 40 cosθ …(i) Squaring both sides, we get ⇒ 81sin2 θ = 1681+1600 cos2 θ – 2(41) (40cos θ) [∵ (a – b)2 =a2 +b2 –2ab] ⇒ 81 (1– cos2 θ) =1681+1600 cos2 θ – 3280cosθ ⇒ 81 – 81cos2 θ = 1681 +1600cos2 θ – 3280 cosθ ⇒ 1681cos2 θ –3280cos θ +1600 = 0 ⇒ (41)2 cos2 θ – 2(41) (40cos θ) + (40)2 = 0 ⇒ (41cos θ – 40 )2 = 0