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