Question

Q The numerator of a+√a2-b2/a-√a2-b2 + a-√a2-b2/a+√a2-b2 is

Answer 1

SOLUTION

TO DETERMINE

The numerator of

$\displaystyle \sf{ \frac{a + \sqrt{ {a}^{2} - {b}^{2} } }{a - \sqrt{ {a}^{2} - {b}^{2} }} + \frac{a - \sqrt{ {a}^{2} - {b}^{2} } }{a + \sqrt{ {a}^{2} - {b}^{2} }} }$

EVALUATION

Here the given fraction is

$\displaystyle \sf{ \frac{a + \sqrt{ {a}^{2} - {b}^{2} } }{a - \sqrt{ {a}^{2} - {b}^{2} }} + \frac{a - \sqrt{ {a}^{2} - {b}^{2} } }{a + \sqrt{ {a}^{2} - {b}^{2} }} }$

We first simplify the given fraction

$\displaystyle \sf{ \frac{a + \sqrt{ {a}^{2} - {b}^{2} } }{a - \sqrt{ {a}^{2} - {b}^{2} }} + \frac{a - \sqrt{ {a}^{2} - {b}^{2} } }{a + \sqrt{ {a}^{2} - {b}^{2} }} }$

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

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

$\displaystyle \sf{ = \frac{{2 \bigg[ {a}^{2} + {(\sqrt{ {a}^{2} - {b}^{2} }) }^{2} } \bigg] }{{a}^{2} - ({ {a}^{2} - {b}^{2}) }}}$

$\displaystyle \sf{ = \frac{{2 \bigg[ {a}^{2} + {a}^{2} - {b}^{2} } \bigg] }{{a}^{2} - { {a}^{2} + {b}^{2} }}}$

$\displaystyle \sf{ = \frac{{2 \bigg[ {a}^{2} + {a}^{2} - {b}^{2} } \bigg] }{{a}^{2} - { {a}^{2} + {b}^{2} }}}$

$\displaystyle \sf{ = \frac{{2 \bigg[ 2{a}^{2} - {b}^{2} } \bigg] }{ {b}^{2} }}$

$\displaystyle \sf{ = \frac{{ 4{a}^{2} - 2{b}^{2} } }{ {b}^{2} }}$

Numerator = 4a² - b²

Denominator = b²

FINAL ANSWER

Numerator = 4a² - b²

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