Math, asked by pattabhipmp, 7 months ago

prove that sin⁴A-cos⁴A=2sin²A-1​

Answers

Answered by michellemeel12
1

Step-by-step explanation:

LHS. RHS

Sin⁴A-Cos⁴A. 2Sin²A-1

Bringing -1 from RHS to LHS, we get LHS as

Solving LHS

=Sin⁴A-Cos⁴A+1

= (sin²A)²-(cos²A)²+1

Using Identity a²-b²= (a-b)(a+b)

(sin²A+cos²A)(sin²A-cos²A)+1

We know that

sin²A+cos²A = 1

So we get the result as

sin²A-cos²A+1

We know that

-cos²A+1 = sin²A OR 1 - cos²A = sin²A

Placing Sin²A we get

Sin²A+ sin²A

= 2sin²A

RHS

2Sin²A

So LHS = RHS

Hence Proved.

Identities used

a²-b²= (a-b)(a+b)

sin²A+cos²A = 1

-cos²A+1 = sin²A

Or 1 - cos²A = sin²A

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