Prove that sin4A-cos4A=sin2A-cos2A=2sin2A-1=1-2cos2A

Question : Prove that sin⁴A-cos⁴A=sin²A-cos²A=2sin²A-1=1-2cos²A

Doubt by Jaskirat

Solution : 

sin⁴A-cos⁴A

=(sin²A)² - (cos²A)²

= (Sin²A+cos²A)(sin²A-cos²A)

[∵a²-b²=(a+b)(a-b)]

= (1)(sin²A-cos²A)

[∵sin²θ+cos²θ=1]

= sin²A-cos²A
= sin²A-(1-sin2A)

[∵cos²θ = 1-sin²θ]
=sin²A-1+sin²A
=2sin²A-1

=2(1-cos²A)-1
[∵sin²θ = 1-cos²θ]
=2-2cos²A-1
=1-cos²A