**If 5 sin θ = 3, find the value of (secθ – tanθ)/(secθ + tanθ).**

**Answer :**

Consider ∆ABC be right angled at B and ∠ACB = θ

5 sin θ = 3

sin θ = AB/AC = 3/5

Take AB = 3x then AC = 5x

In right angled ∆ABC

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

⇒ BC^{2} = AC^{2} – AB^{2}

BC^{2} = (5x)^{2} – (3x)^{2}

BC^{2} = 25x^{2} – 9x^{2} = 16x^{2}

⇒ BC^{2} = (4x)^{2}

⇒ BC = 4x

In right angled ∆ABC

sec θ = hypotenuse/base

⇒ sec θ = AC/BC = 5x/4x = 5/4

In right angled ∆ABC

tan θ = perpendicular/base

⇒ tan θ = AB/BC = 3x/4x = ¾

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