Question

Sin4A+cos4
A+2sin2A cos 2A

Answer 1

LHS = sin^4A + cos^4A + 2sin^2A.cos^2A

= (sin^2A)^2+(cos^2A)^2+2sin^2A.cos^2A =[(sin^2A+cos^2A)^2−2sin^2A.cos^2A]+2sin^2A.cos^2A

°.° [a^2+b^2=(a+b)^2−2ab]

=[1^2−2sin^2A cos^2A]+2sin^2A.cos^2A °.°(sin^2A+cos^2A=1)

=1−2 sin^2A cos^2A+2sin^2A.cos^2A

= 1-0

=1