Question

31. (a) If 9 sin 0+40 cos 0 = 41, prove that 41 cos 0 = 40.​

Answer 1

Answer:

Step-by-step explanation:

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