# If 9 sin θ + 40 cos θ= 41, find the value of cos θ and cosec θ

Given: 9 sin θ + 40 cos θ= 41

9sinθ = 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

Now, putting the value of cos θ in Eq. (i), we get

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