Question

Q. 2 (a) If a+c² ab+cd =ab+cd/b seqaure + d seqaure
Then prove that a b=cd​

Answer 1

Correct Question:-

If (a² + c²) / ab + cd = (ab + cd) / b² + d² , then prove that a/b = c/d.

Answer:-

Given:-

(a² + c²)/ab + cd = (ab + cd)/b² + d²

On cross multiplication we get,

⟹ (a² + c²)(b² + d²) = (ab + cd)²

Using (a + b)² = a² + b² + 2ab in LHS we get,

⟹ a²(b² + d²) + c²(b² + d²) = (ab)² + (cd)² + 2(ab)(cd)

⟹ a²b² + a²d² + b²c² + c²d² = a²b² + c²d² + 2abcd

[ a²b² , c²d² get cancelled both sides ]

⟹ a²d² + b²c² = 2abcd

⟹ (ad)² + (bc)² - 2(ab)(cd) = 0

using a² + b² - 2ab = (a - b)² we get,

⟹ (ad - bc)² = 0

⟹ ad - bc = √0

⟹ ad = bc

Dividing both sides by bd we get,

⟹ ad/bd = bc/bd

⟹ a/b = c/d

Hence, Proved.

Answer 2

Right Question:-

If (a² + c²) / ab + cd = (ab + cd) / b² + d² , then prove that a/b = c/d.

Given:-

(a² + c²)/ab + cd = (ab + cd)/b² + d²

On cross multiplication we get,

★ (a² + c²)(b² + d²) = (ab + cd)²

Using (a + b)² = a² + b² + 2ab in LHS we get,

★ a²(b² + d²) + c²(b² + d²) = (ab)² + (cd)² + 2(ab)(cd)

★ a²b² + a²d² + b²c² + c²d² = a²b² + c²d² + 2abcd

[ a²b² , c²d² get cancelled both sides ]

★ a²d² + b²c² = 2abcd

★ (ad)² + (bc)² - 2(ab)(cd) = 0

using a² + b² - 2ab = (a - b)² we get :-

★ (ad - bc)² = 0

★ ad - bc = √0

★ ad = bc

Dividing both sides by bd we get,

★ ad/bd = bc/bd

★ a/b = c/d

★|★Hence Proved :)

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