Dividing $(a + b)^{2} + (a - b)^{2}$ by $(a^{2} + b^{2})$ gives ___

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Solution

Given that numerator is $(a + b)^{2} + (a - b)^{2}$ = $a^{2} + b^{2} + 2 a b) + (a^{2} + b^{2} - 2 a b)$
= $2 (a^{2} + b^{2})$

Hence $(a + b)^{2} + (a - b)^{2}$ by $(a^{2} + b^{2})$ = $\frac{2 (a^{2} + b^{2})}{(a^{2} + b^{2})}$
= 2

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Dividing $(a + b)^{2} + (a - b)^{2}$ by $a^{2} + b^{2}$ gives:

Q. Simplify

\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}}}

\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}}}

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

Q. The denominator of

\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}}}

is ____.

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Dividing (a+b)2+(a−b)2 by (a2+b2) gives ___

Given that numerator is (a+b)2+(a−b)2 = a2+b2+2ab)+(a2+b2−2ab) = 2(a2+b2) Hence (a+b)2+(a−b)2 by (a2+b2) = 2(a2+b2)(a2+b2) = 2

Dividing $(a + b)^{2} + (a - b)^{2}$ by $(a^{2} + b^{2})$ gives ___

Given that numerator is $(a + b)^{2} + (a - b)^{2}$ = $a^{2} + b^{2} + 2 a b) + (a^{2} + b^{2} - 2 a b)$
= $2 (a^{2} + b^{2})$

Hence $(a + b)^{2} + (a - b)^{2}$ by $(a^{2} + b^{2})$ = $\frac{2 (a^{2} + b^{2})}{(a^{2} + b^{2})}$
= 2