Math, asked by rajuvarghese299, 5 months ago

sin⁶A-cos⁶A = 1-3 sin²A.cos²A

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Answered by ravindras931

4

Step-by-step explanation:

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Answered by ansarahmadansari68

2

Answer:

sin^6A + cos^6A

= (sin^2A)3 + (cos^2A)3

= (sin^2A + cos^2A)(sin^4A – sin^2Acos^2A + cos^4A)

= 1 (sin^4A – sin^2Acos^2A + cos^4A + 2sin^2Acos^2A – 2sin^2Acos^2A)

= (sin2A + cos2A) – 3sin2Acos2A

= 1 – 3sin^2Acos^2A

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