If $P=\frac{(\sqrt{7}-\sqrt{6})}{(\sqrt{7}+\sqrt{6})}$, then what is the value of $\left(\mathrm{P}+\frac{1}{\mathrm{P}}\right) ?$

$P=\frac{\sqrt{7}-\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{(\sqrt{7}-\sqrt{6})(\sqrt{7}-\sqrt{6})}{(\sqrt{7}+\sqrt{6})(\sqrt{7}-\sqrt{6})}$ (On rationalising the denominator) $=\frac{(\sqrt{7}-\sqrt{6})^{2}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}$ [ (a+b) (a–b) = a²–b²] $=\frac{7+6-2 \sqrt{7 \times 6}}{7-6}=13-2 \sqrt{42}$ $\therefore \frac{1}{\mathrm{P}}=\frac{1}{13-2 \sqrt{42}}=\frac{13+2 \sqrt{42}}{(13-2 \sqrt{42})(13+2 \sqrt{42})}$ $=\frac{13+2 \sqrt{42}}{169-168}=13+2 \sqrt{42}$ $\therefore \mathrm{P}+\frac{1}{\mathrm{P}}=13-2 \sqrt{42}+13+2 \sqrt{42}=26$

$P=\frac{\sqrt{7}-\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

$\therefore \mathrm{P}+\frac{1}{\mathrm{P}}=\frac{\sqrt{7}-\sqrt{6}}{\sqrt{7}+\sqrt{6}}+\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}-\sqrt{6}}$

$=\frac{(\sqrt{7}-\sqrt{6})^{2}+(\sqrt{7}+\sqrt{6})^{2}}{(\sqrt{7}+\sqrt{6})(\sqrt{7}-\sqrt{6})}=\frac{2(7+6)}{7-6}=2 \times 13=26$

[ (a + b)² + (a – b)² = 2 (a² + b²)]