Question

Prove that: sin4 A + cos⁴ A = 1 – 2sin² A cos²A​

Answer 1

Step-by-step explanation:

We have,

LHS = sin 4 A - cos 4 A

⇒ LHS = (sin 2 A) 2 −(cos 2 A) 2

⇒ LHS = (sin 2 A+cos 2 A)(sin 2 A−cos 2 A)

⇒ LHS = sin 2 A −cos 2 A [∵sin 2 A+cos 2 A=1]

⇒ LHS = sin 2 A −(1−sin 2 A ) = 2sin 2 A−1

⇒ LHS = 2(1−cos 2A)−1=1−2cos 2

A = RHS

Answer 2

To prove :

sin⁴ A + cos⁴ A = 1 – 2sin² A cos²A

Proof :

LHS :

sin⁴ A + cos ⁴ A

=> [ sin² A ]² + [ cos ² A ]²

Let , sin ²A = k , and cos² A = l

=> k² + l²

=> k² + l² + 2kl - 2kl

=> ( k + l )² - 2kl

Note : Instead of using ( a + b)² algebraic identity , we could have used ( a - b)² but didn't because of the property that , sin² A + cos²A = 1 or k + l = 1 .

=> [ sin² A ] + [ cos² A ]² - 2 [ sin²A ][ cos² A ]

=> 1 - 2 sin²A cos ² A

Hence Proved

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Additional Information :

• sin² A + cos² A = 1

• tan² A + 1 = sec ² A

• cot² A + 1 = cosec² A

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