Prove that: (sin⁴θ – cos⁴θ + 1) cosec²θ = 2

Formula: -

(i)

(ii) a² – b² = (a + b)(a – b)

L.H.S =

(sin⁴θ – cos⁴θ + 1) cosec²θ = ((sin²θ)² – (cos²θ)² + 1)cosec²θ

using formula (ii)

⇒ (sin⁴θ – cos⁴θ + 1) cosec²θ = ((sin²θ + cos²θ)(sin²θ – cos²θ) + 1)cosec²θ

⇒ (sin⁴θ – cos⁴θ + 1)cosec²θ = (sin²θ – cos²θ + 1)cosec²θ

⇒ (sin⁴θ – cos⁴θ + 1)cosec²θ = (sin²θ + (1 – cos²θ))cosec²θ

using formula(i)

⇒ (sin⁴θ – cos⁴θ + 1)cosec²θ = (sin²θ + sin²θ)cosec²θ

⇒ (sin⁴θ – cos⁴θ + 1)cosec²θ = 2sin²θ.cosec²θ

⇒ (sin⁴θ – cos⁴θ + 1)cosec²θ = 2

L.H.S = R.H.S

Hence, Proved.