Prove the following identities:

`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`

`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`

`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`

#### Solution

(i) We have,

`LHS = cos4^4 A – cos^2 A = cos^2 A (cos^2 A – 1)`

`= – cos^2 A (1 – cos^2 A) = – cos^2 A sin2A`

`= –(1 – sin^2 A) sin^2 A = – sin^2 A + sin^4 A`

`= sin^4 A – sin^2 A = RHS`

(ii) We have,

`LHS = cot^4 A – 1 = (cosec^2 A – 1)^2 – 1`

`[∵ cot^2 A = cosec^2 A –1 ⇒ cot^4 A = (cosec^2 A – 1)^2 ]`

`= cosec^4 A – 2 cosec^2 A + 1 – 1`

`= cosec^4 A – 2cosec^2 A = RHS`

(iii) We have,

`LHS = sin^6 A + cos^6 A = (sin^2 A)^3 + (cos2 A)^3`

`= (sin^2 A + cos^2 A) {(sin^2 A)^2 + (cos^2 A)^2 – sin^2 A cos^2 A)}`

`[∵ a^3 + b^3 = (a + b) (a^2 – ab + b^2 )]`

`={(sin^2 A)^2 + (cos^2 A)^2 + 2 sin^2 A cos^2 A – sin^2 A cos^2 A}`

`= [(sin^2 A + cos^2 A)^2 – 3 sin^2 A cos^2 A]`

`= 1 – 3sin^2 A cos^2 A = RHS`