Question Number 180837 by Mastermind last updated on 17/Nov/22
$$\mathrm{Without}\:\mathrm{using}\:\mathrm{table},\:\mathrm{find}\:\mathrm{the}\:\mathrm{values} \\ $$$$\mathrm{of}: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Nov/22
$$\frac{\mathrm{1}}{\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\frac{\left\{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)−\left(\mathrm{1}−\sqrt{\mathrm{3}}\:\right)\right\}\left\{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)+\left(\mathrm{1}−\sqrt{\mathrm{3}}\:\right)\right\}}{\left\{\left(\mathrm{1}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} \right\}^{\mathrm{2}} } \\ $$$$\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:−\mathrm{1}+\sqrt{\mathrm{3}}\:\right)\left(\mathrm{1}+\sqrt{\mathrm{3}}\:+\mathrm{1}−\sqrt{\mathrm{3}}\:\right)}{\left(\mathrm{1}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right)\left(\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{3}}\: \\ $$
Answered by Socracious last updated on 19/Nov/22
$$\:\:\:\:\:\:\sqrt{\mathrm{3}} \\ $$