Question

(ab+cd)^2 = (a^2+c^2) (b^2+d^2) prove that a/b = c/d​

Answer 1

Answer:

Step-by-step explanation:

(ab+cd)^2 = (a^2+c^2) (b^2+d^2)

=> a^2b^2 + 2abcd + c^2d^2 = a^2b^2 + b^2c^2 + a^2d^2 + c^2d^2

=> 2abcd = b^2c^2 + a^2d^2

=> b^2c^2 + a^2d^2 - 2abcd  = 0

=> (bc - ad)^2 = 0

=> bc - ad = 0

=> bc = ad

=> a/b = c/d ----divide by bd

Answer 2

Answer:

Proved below.

Step-by-step explanation:

Here,

       a / b = c / d

       ⇒ ad = cb        ...( 1 )

       ⇒ ( ad )^2 = ( cb )^2   ..( 2 )

Now,

⇒ ( ab + cd )^2

⇒ ( ab )^2 + ( cd )^2 + 2( ab )( cd )     { Using ( a + b )^2 = a^2 + b^2 + 2ab }

⇒ ( ab )^2 + ( cd )^2 + 2( abcd )

⇒ ( ab )^2 + ( cd )^2 + 2( ad )( cb )

⇒ ( ab )^2 + ( cd )^2 + 2( cb )( cb )          { ad = cb }

⇒ ( ab )^2 + ( cd )^2 + 2( cb )^2

⇒ ( ab )^2 + ( cd )^2 + ( cb )^2 + ( cb )^2

⇒ ( ab )^2 + ( cb )^2 + ( cd )^2 + ( ad )^2

⇒ b^2( a^2 + c^2 ) + d^2( c^2 + a^2 )         { cb = ad }

⇒ ( a^2 + c^2 )( b^2 + d^2 )

Hence,

    ( ab + cd )^2 = ( a^2 + c^2 )( b^2 + d^2 )