Math, asked by reddy337, 1 month ago

Correct Answer pls....

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Answers

Answered by Mathkeeper
1

Step-by-step explanation:

We have,

 \frac{ \sqrt{ {a}^{2} -  {b}^{2}  }  + a}{ \sqrt{ {a}^{2}  +  {b}^{2} }  + b}  \div  \frac{ \sqrt{ {a}^{2} +  {b}^{2}  }  - b}{a -  \sqrt{ {a}^{2} -  {b}^{2}  } }  \\

 =  \frac{ \sqrt{ {a}^{2} -  {b}^{2}  }  + a}{ \sqrt{ {a}^{2}  +  {b}^{2} }  + b}   \times   \frac{a -  \sqrt{ {a}^{2}  -   {b}^{2}  } }{\sqrt{ {a}^{2}  +  {b}^{2}  }  - b}  \\

 =  \frac{  \{\sqrt{ {a}^{2} -  {b}^{2}  }  + a \} \{ a -  \sqrt{ {a}^{2}  -   {b}^{2}  } \}}{  \{\sqrt{ {a}^{2}  +  {b}^{2} }  + b \} \{ \sqrt{ {a}^{2}  +  {b}^{2}  }  - b \} }     \\

 =  \frac{  (a) ^{2} -  (\sqrt{ {a}^{2}  -   {b}^{2}  })^{2}  }{ (\sqrt{ {a}^{2}  +  {b}^{2}  }) ^{2}   - (b)^{2} }     \\

 =  \frac{  a^{2} -  {a}^{2}   +    {b}^{2}   }{  {a}^{2}  +  {b}^{2}     - b^{2} }     \\

 =  \frac{   {b}^{2}   }{  {a}^{2}  }     \\

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