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प्रश्न
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.
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उत्तर
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.
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