solve both the question Q12...
first answer will be Mark as brainliest.....
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( sin⁴A + cos⁴A) / 1 - 2 sin²A cos²A = 1
LHS = ( sin⁴A + cos⁴A) / 1 - 2 sin²A cos²A
= [ (sin²A)² + (cos²A)²] / 1 - 2 sin²A cos²A
= ( sin²A + cos²A)² - 2 sin²A cos²A / 1 - 2 sin²A cos²A
= (1)² - 2 sin²A cos²A / 1 - 2 sin²A cos²A
= 1 - 2 sin²A cos²A / 1 - 2 sin²A cos²A
= 1 = RHS
LHS = ( sin⁴A + cos⁴A) / 1 - 2 sin²A cos²A
= [ (sin²A)² + (cos²A)²] / 1 - 2 sin²A cos²A
= ( sin²A + cos²A)² - 2 sin²A cos²A / 1 - 2 sin²A cos²A
= (1)² - 2 sin²A cos²A / 1 - 2 sin²A cos²A
= 1 - 2 sin²A cos²A / 1 - 2 sin²A cos²A
= 1 = RHS
Anonymous:
thanks a lot mate
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