Question

pls solve this question ​

Answer 1

Answer:

Option(3)

Step-by-step explanation:

Given: cos⁶A + sin⁶A

= (cos²A)³ + (sin²A)³

∴ a³ + b³ = (a+b)(a²-ab+b²)

= (cos²A+sin²A)(cos⁴A+sin⁴A-cos²Asin²A)

= (1)(cos⁴A+sin⁴A+2cos²Asin²A-2cos²Asin²A-cos²Asin²A)

= [(cos²A+sin²A)² - 2cos²Asin²A - cos²Asin²A]

= 1 - 3cos²Asin²A

= 1 - (3/4) * 4cos²Asin²A

= 1 - (3/4) * (2cosAsinA)²

= 1 - (3/4) * (sin2A)²

= 1 - (3/4) * sin²(2A)

On comparing with 1 - k sin²(2A), we get

k = (3/4)

Hope it helps!

Answer 2

Answer:

Step-by-step explanation:

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