sin*4 A-cos*4 A/ sin*2A-cos*2A=1
Answers
Answered by
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
••••
=[(sin²A)² - (cos²A)²]/(sin²A -cos²A )
=[(sin²A+cos²A)(sin²A-cos²A)]/(sin²A-cos²A)
after cancellation , we get
= sin²A + cos²A
= 1 [ By trigonometric identity]
= RHS
••••
Similar questions