Use the small angle approximation on cos θ and sin θ

cos θ ≈ 1 - ½θ²
sin θ ≈ θ

cos θ ≈ 1 - ½θ²
sin θ ≈ θ

So the denominator √3 cos θ - sin θ

≈ √3 (1 - ½θ²) - θ


BC ≈ √3 / (√3 (1 - ½θ²) - θ)

BC ≈ 1 / ( 1 - ½θ² - θ/√3)

Rationalising the denominator,

BC ≈ 1 / (1 - ½θ² - θ/√3) × (1 - ½θ² + θ/√3) / (1 - ½θ² + θ/√3)

BC ≈ (1 - ½θ² + θ/√3) / ((1 - ½θ²)² - (θ/√3)²)

BC ≈ (1 + θ/√3 - ½θ²) / ((1 - ½θ²)² - (θ/√3)²)

BC ≈ (1 + θ/√3 - ½θ²) / (1 - θ² + ¼θ⁴ - θ²/3)

BC ≈ (1 + θ/√3 - ½θ²) / (1 + ¼θ⁴ - 4/3 θ²)

Given that θ is sufficiently small,
1 ≫ ¼θ⁴ - 4/3 θ²

So, 1 + ¼θ⁴ - 4/3 θ² ≈ 1

BC ≈ (1 + θ/√3 - ½θ²) / 1

BC ≈ 1 + θ/√3 - ½θ²

p = -½

Alternative explanation :


BC = √3 / (√3 cos θ - sin θ)

BC = 1 / (cos θ - sinθ / √3)

1/BC = cos θ - sinθ / √3

Since AC = 1 , then AC/BC = cos θ - sinθ / √3

Given that θ is sufficiently small, we can use the small angle approximation.

Furthermore, as θ becomes very small and closer to 0,

we notice that while ∠BAC is fixed, ∠ABC actually gets bigger and closer to π/3.

This also means that both the lengths of AC and BC are getting shorter.

The length of AB is approaching 0 and the length of BC is approaching AC (getting closer to it)

So we can say that when θ is sufficiently small,

BC ≈ AC = 1
BC ≈ 1

Using cosθ ≈ 1 - ½θ² and sin θ ≈ θ,

1/BC ≈ 1 - ½θ² - θ/√3

Since BC ≈ 1, then 1/BC ≈ 1. This also means that 1/BC ≈ BC

∴ BC ≈ 1/BC ≈ 1 - ½θ² - θ/√3