sin4A – cos4A = 1 – 2cos2A हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा. कृती: डावी बाजू = `square` = (sin2A + cos2A) `(square)` = `1 (square)` .....`[sin^2"A" + square = 1]` = `square` – cos2A .....[sin2A = 1 – cos2A] = `square` = उजवी बाजू

डावी बाजू = sin4A – cos4A = (sin2A)2 – (cos2A)2 = (sin2A + cos2A) (sin2A – cos2A) .....[∵ a2 – b2 = (a + b)(a – b)] = 1(sin2A – cos2A) .....[∵ sin2A + cos2A = 1] = sin2A – cos2A = 1 – cos2A – cos2A .....[sin2A = 1 – cos2A] = 1 – 2cos2A = उजवी बाजू

Sin4A – cos4A = 1 – 2cos2A हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा. कृती: डावी बाजू = □ = (sin2A + cos2A) (□) = 1(□) .....[sin2A+□=1] = □ – cos2A .....[sin2A = 1 – cos2A] = □ = उजवी बाजू

Question

sin⁴A – cos⁴A = 1 – 2cos²A हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.

कृती: डावी बाजू = `square`

= (sin²A + cos²A) `(square)`

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

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

= `square`

= उजवी बाजू

Sum

Solution

डावी बाजू = sin⁴A – cos⁴A

= (sin²A)² – (cos²A)²

= (sin²A + cos²A) (sin²A – cos²A) .....[∵ a² – b² = (a + b)(a – b)]

= 1(sin²A – cos²A) .....[∵ sin²A + cos²A = 1]

= sin²A – cos²A

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

= 1 – 2cos²A

= उजवी बाजू

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