If $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=a+b \sqrt{35},$ find the value of $a$ and $b$.

(A) To solve $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=a+b \sqrt{35}$,we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{7}+\sqrt{5})$.
$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} \times \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} = \frac{(\sqrt{7}+\sqrt{5})^2}{(\sqrt{7})^2-(\sqrt{5})^2}$
$= \frac{7+5+2\sqrt{35}}{7-5}$
$= \frac{12+2\sqrt{35}}{2}$
$= 6+\sqrt{35}$
Comparing $6+\sqrt{35}$ with $a+b\sqrt{35}$,we get $a=6$ and $b=1$.