If 3 cot θ = 4, find the value of (5 sinθ – 3 cosθ)/(5 sinθ + 3 cosθ).
Answer :
3 cot θ = 4
⇒ cot θ = 4/3
Consider ∆ABC be right angled at B and ∠ACB = θ
cot θ = BC/AB = 4/3
Take BC = 4x then AB = 3x
In right angled ∆ABC
AC2 = AB2 + BC2
AC2 = (3x)2 + (4x)2
⇒ AC2 = 9x2 + 16x2 = 25x2
AC2 = (5x)2
⇒ AC = 5x
In right angled ∆ABC
sin θ = perpendicular/hypotenuse
⇒ sin θ = AB/AC
sin θ = 3x/5x = 3/5
In right angled ∆ABC
cos θ = base/hypotenuse
⇒ cos θ = BC/AC
cos θ = 4x/5x
= 4/5
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