Prove the following identities :

sin⁴ A – cos⁴ A = 2 sin² A – 1 = 1 – 2 cos² A = sin² A – cos² A

Given:

Taking I term

= sin⁴ A – cos⁴ A → I term

= (sin² A)² – (cos² A)²

= (sin² A – cos² A)(sin² A+ cos² A )

[∵ (a² – b²) = (a + b) (a – b)]

= (sin² A – cos² A)(1) [∵ cos² θ + sin² θ = 1]

= (sin² A – cos² A) …(i) → IV term

From Eq. (i)

= {sin² A – (1 – sin² A)} [∵ cos² θ + sin² θ = 1]

= sin² A – 1 + sin² A

= 2 sin² A – 1 → II term

Again, From Eq. (i)

= {(1 – cos² A) – cos² A } [∵ cos² θ + sin² θ = 1]

=1 – 2 cos² A → III term

Hence, I = II = III = IV

Hence Proved