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
AC2 = AB2 + BC2
⇒ BC2 = AC2 – AB2
BC2 = (5x)2 – (3x)2
BC2 = 25x2 – 9x2 = 16x2
⇒ BC2 = (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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