If cosec θ – sin θ = l; sec θ – cos θ = m, then prove that l² m² (l² + m² + 3) = 1

Formula: - (i) a³ + b³ = (a + b)³ – 3 a b (a + b)

(ii) Sin²θ + Cos²θ = 1

Given : -

Cosecθ - sinθ = l

Secθ – Cosθ = m

We have to prove that : l² m² (l² + m² + 3) = 1

Cosecθ – sinθ = l

And secθ – Cosθ = m

putting (a)and(b)

Multiplying with

= (cos²θ)³ + ( sin²θ)³ + 3sin²θCos²θ

l² m² (l² + m² + 3) = 1