Question Number 192733 by Mingma last updated on 25/May/23
Answered by MM42 last updated on 25/May/23
$$\frac{\mathrm{1}−\frac{\mathrm{2}{tanA}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}}{\mathrm{1}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}}={tanA}−\mathrm{1} \\ $$$$\frac{{tan}^{\mathrm{2}} {A}−\mathrm{2}{tanA}+\mathrm{1}}{\mathrm{2}}={tanA}−\mathrm{1} \\ $$$${tan}^{\mathrm{2}} {A}−\mathrm{4}{tanA}+\mathrm{3}=\mathrm{0} \\ $$$${tanA}=\mathrm{1}\:×\: \\ $$$${or}\:\:{tanA}=\mathrm{3}\:\checkmark\:\Rightarrow{tan}^{\mathrm{2}} {A}=\:\:\mathrm{9} \\ $$$$ \\ $$
Commented by Mingma last updated on 25/May/23
Perfect ��