Math, asked by tanya374, 1 year ago

sin*4 A-cos*4 A/ sin*2A-cos*2A=1

Answers

Answered by mysticd
1
LHS = (sin⁴A-cos⁴A)/(sin²A-cos²A)

=[(sin²A)² - (cos²A)²]/(sin²A -cos²A )

=[(sin²A+cos²A)(sin²A-cos²A)]/(sin²A-cos²A)

after cancellation , we get

= sin²A + cos²A

= 1 [ By trigonometric identity]

= RHS

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