Question

Show that (√7+i√3/√7-i√3 + √7-i√3/√7+i√3) is real​

Answer 1

Explanation:

Answer 2

Explanation:

`\large\underline{\sf{Solution-}}`

Consider,

`\rm \longmapsto\:( √(7) + i √(3) )/( √(7) - i √(3)) + ( √(7) - i √(3) )/( √(7) + i √(3) )`

On taking LCM, we get

`\rm \:  =  \: \frac{ {( √(7) + i √(3))}^(2) + {( √(7) - i √(3))}^(2) }{( √(7) + i √(3))( √(7) - i √(3))}`

We know,

`\purple{\rm \longmapsto\:\boxed{\tt{ {(x + y)}^(2) + {(x - y)}^(2) = 2( {x}^(2) + {y}^(2))}}} \\`

and

`\purple{\rm \longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^(2) - {y}^(2) \: }}} \\`

So, using these Identities, we get

`\rm \:  =  \: \frac{2\bigg[ {( √(7))}^(2) + {(i √(3)) }^(2) \bigg]}{ {( √(7)) }^(2) - {(i √(3))}^(2) }`

`\rm \:  =  \: \frac{2(7 + 3 {i}^(2))}{7 - {3i}^(2) }`

We know,

`\purple{\rm \longmapsto\:\boxed{\tt{ {i}^(2) = - 1}}} \\`

So, using this, we get

`\rm \:  =  \: (2(7 - 3))/(7 + 3)`

`\rm \:  =  \: (2 * 4)/(10)`

`\rm \:  =  \: (4)/(5)`

Hence,

`\red{\rm\implies \:\boxed{\sf{ ( √(7) + i √(3) )/( √(7) - i √(3)) + ( √(7) - i √(3) )/( √(7) + i √(3) ) \: is \: purely \: real}}}`