Math, asked by kotharishm, 3 months ago

11. What is the value of sin⁴O-cos⁴O
sin²O-cos²O

Answers

Answered by Anonymous
5

Answer:

We, have

LHS = sin

4

A + cos

4

A

⇒ LHS = (sin

2

A)

2

+(cos

2

A)

2

+2sin

2

Acos

2

A−2sin

2

Acos

2

A [Adding and subtracting 2 sin

2

A cos

2

A ]

⇒ LHS = (sin

2

A+cos

2

A)

2

−2sin

2

Acos

2

A=1−2sin

2

Acos

2

A=RHS

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