Question

prove that sin⁴A-cos⁴A=2sin²A-1​

Answer 1

This is the proper solution for upper equation.

Answer 2

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