Question

Sec^4A-cos^4A=1-2cos^2A

Answer 1

Oops , your question is wrong ! Pz write correct question .
Correct question is ---> sin⁴A - cos⁴A = 1 - 2cos²A

LHS = sin⁴A - cos⁴A
= (sin²A - cos²A)(sin²A + cos²A)

As you know, sin²x + cos²x = 1
So, sin²A + cos²A = 1

∴ LHS = (sin²A - cos²A) × 1
= (1 - cos²A - cos²A) [ sin²x = 1 - cos²x from identities ]
= 1 - 2cos²A = RHS
[Hence, proved]

Answer 2

Hey friend,

Here is your answer,
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To prove  :  sin⁴A - cos⁴A = 1 - 2cos²A 

LHS:-

=> sin⁴A - cos⁴A 

=> (sin²A - cos²A)(sin²A + cos²A) 

=> (sin²A - cos²A) × 1                     {sin²x + cos²x = 1}

=> (1 - cos²A - cos²A)                    {sin²x  = 1 - cos²x}

=> 1 - 2cos²A = RHS 

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