**If tan θ = p/q, find the value of (p sin θ – q cos θ)/ (p sin θ + q cos θ).**

**Answer :**

tan θ = p/q

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

tan θ = BC/AB = p/q

BC = px then AB = qx

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

AC^{2} = (px)^{2} + (qx)^{2}

⇒ AC^{2} = p^{2}x^{2} + q^{2}x^{2}

⇒ AC^{2} = x^{2} (p^{2} + q^{2})

AC = √x^{2} (p^{2} + q^{2})

⇒ AC = x(√p^{2} + q^{2})

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = BC/AC

sin θ = px/x(√p^{2} + q^{2})

sin θ = p/(√p^{2} + q^{2})

In right angled ∆ABC

cos θ = base/hypotenuse

⇒ cos θ = AB/AC

cos θ = qx/x(√p^{2} + q^{2})

cos θ = q/(√p^{2} + q^{2})

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