**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

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (3x)^{2} + (4x)^{2}

⇒ AC^{2} = 9x^{2} + 16x^{2} = 25x^{2}

AC^{2} = (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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