Question

prove that sin²*²A-cos²*²A=sin²A - cos²A​

Answer 1

Prove that cos⁴A-cos²A=sin⁴A-sin²A.

LHS = cos⁴ A - cos² A

= cos²A( cos² A - 1 )

= ( 1 - sin² A )[ -( 1 - cos² A ) ]

= ( 1 - sin² A ) ( - sin² A )

= - sin² A + sin⁴ A

= sin⁴ A - sin² A

= RHS

hope this answer helpful

Answer 2

sin^4 A -cos^4 A

= (sin^2 A - cos^A)*(sin^2 A + cos^2 A)

= (sin^2 A - cos^A)*1

= (sin^2 A - cos^A)

=R.H.S.

Answer 3

(Sin^2A)^2-(Cos^2A)^2

=Sin^2A+Cos^2A)(Sin^2A-Cos^2A)

=1×(Sin^2A-Cos^2)

=(Sin^2A-Cos^2A) Ans