Question

plz solve 13 ques......​

Answer 1

Step-by-step explanation:

(sin6A + cos6A) - 3(sin4A+ cos4A) +1=0

1.We know the algebraic identity:

a³+b³+3ab(a+b) = (a+b)³

=> a³+b³ = (a+b)³-3ab(a+b)

By Trigonometric identity:

sin²A + cos²A = 1

(sin6A + cos6A)= (sin²)³+(cos²)³

=(sin²A+cos²)³-3sin²Acos²A(sin²Acos²A)

= 1-3sin²Acos²A ----(1)

2.we know the algebraic identity

a²+b²+2ab = (a+b)²

=> a²+b² = (a+b)²-2ab

here,

sin⁴A + cos⁴A

= (sin²A)²+(cos²A)²

= (sin²A+cos²A)²-2sin²Acos²A

= 1 - 2sin²Acos²A ----(2)

Now ,

LHS

(sin6A + cos6A) - 3(sin4A+ cos4A) +1

from (1)&(2),we get

= 2(1-3sin²Acos²A)-3(1-2sin²Acos²A)+1

= 2-6sin²Acos²A-3-6sin²Acos²A+1

= 2-3+1

= 0

= RHS

Answer 2

Answer

In Attachment ..