Question

sin^6A+cos^6A+3sin^A.cos^2A+4=k, then find the value of k

Answer 1

Step-by-step explanation:

follow the above steps

Answer 2

Answer:

sin^6A+cos^6A+3sin²A+cos²A+4=k

➡(sin²A)³+(cos²A)³+3sin²Acos²A+4=k

➡(sin⁴A+cos⁴A-sin²Acos²A)+3sin²Acos²A+4=k

➡{(sin²A+cos²A)²-2sin²Acos²A-sin²Acos²A}+3sin²Acos²A+4=k

➡1-3sin²Acos²A+3sin²Acos²A+4=k

➡k=1+4=5