Question

sin⁴A – cos⁴A = 1 – 2cos²A. For proof of this complete the activity given below.

Activity:

L.H.S. = `square`

 = (sin²A + cos²A) `(square)`

= `1 (square)`   ...`[sin^2"A" + square = 1]`

= `square` – cos²A   ...[sin²A = 1 – cos²A]

= `square`

= R.H.S.

Answer 1

L.H.S. = \[\boxed{\text{sin}^4A - \text{cos}^4A}\] 

= (sin²A)² – (cos²A)²

 = \[{(\text{sin}^2A + \text{cos}^2A) (\boxed{\text{sin}^2A - \text{cos}^2A})}\]   ...[∵ a² – b² = (a + b)(a – b)]

= \[1(\boxed{\text{sin}^2A - \text{cos}^2A})\]   ...[∵ sin²A + \[\boxed{\text{cos}^2\text{A}}\] = 1]

= sin²A – cos²A

= \[\boxed{1 - \text{cos}^2A} - \text{cos}^2A\]   ...[sin²A = 1 – cos²A]

= \[\boxed{1 - 2\text{cos}^2A}\]

= R.H.S.