Find tanθ – cotθ in terms of p and q.

If q cosθ = p, find tanθ – cotθ in terms of p and q.

Answer :

Consider ABC as a triangle right angled at B and ∠ACB = θ

q cos θ = p

cos θ = BC/AC = p/q

Take BC = px then AC = qx

In right angled ∆ABC

AC² = AB² + BC²

AB² = AC² – BC²

AB² = (qx)² – (px)²

⇒ AB² = q²x² – p²x²

AB² = (q² – p²)x²

AB = √(q² – p²) x²

⇒ AB = (√q² – p²)x

In right angled ∆ABC

tan θ = perpendicular/base

tan θ = AB/BC = [(√q² – p²)x]/px

⇒ tan θ = (√q² – p²)/p

In right angled ∆ABC

cot θ = base/perpendicular

cot θ = BC/AB = px/[(√q² – p²)x]

⇒ cot [(√q² – p²)x] = p/(√q² – p²)

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