Question Number 140803 by jlewis last updated on 12/May/21
Answered by Ar Brandon last updated on 12/May/21
$$\mathrm{tany}=\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:,\:\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{y}=\mathrm{sec}^{\mathrm{2}} \mathrm{y} \\ $$$$\Rightarrow\mathrm{cosy}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{y}}}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} }} \\ $$