pls solve this question
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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!
hariharan108:
sir I didn't understand 4th step
Awesome answer, annaya❤
Thank you chelli
Could you please write the 4th step here?
Superrrr answer my dear
thabks dear
Answered by
0
Answer:
Step-by-step explanation:
BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST
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