Question Number 21241 by tawa tawa last updated on 17/Sep/17
$$\mathrm{Find}\:\mathrm{y}\:\mathrm{in}\:\mathrm{3rd}\:\mathrm{quadrant} \\ $$$$\mathrm{tan}\left(\mathrm{y}\:−\:\mathrm{30}\right)\:=\:\mathrm{cot}\left(\mathrm{y}\right) \\ $$
Answered by sma3l2996 last updated on 17/Sep/17
$${tan}\left({y}−\mathrm{30}\right)={cot}\left({y}\right) \\ $$$${tan}\left({y}−\mathrm{30}\right)=\frac{\mathrm{1}}{{tan}\left({y}\right)}\Leftrightarrow\frac{{tan}\left({y}\right)−{tan}\left(\mathrm{30}\right)}{\mathrm{1}+{tan}\left({y}\right){tan}\left(\mathrm{30}\right)}=\frac{\mathrm{1}}{{tany}} \\ $$$${with}\:\:{tan}\left(\mathrm{30}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${tan}\left({y}\right)\left({tan}\left({y}\right)−{tan}\left(\mathrm{30}\right)\right)=\mathrm{1}+{tan}\left({y}\right){tan}\left(\mathrm{30}\right) \\ $$$${tan}^{\mathrm{2}} \left({y}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{tan}\left({y}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{tan}\left({y}\right)=\mathrm{1} \\ $$$${tan}^{\mathrm{2}} \left({y}\right)−\mathrm{2}\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{tan}\left({y}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{4}=\frac{\mathrm{16}}{\mathrm{3}}=\left(\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${tan}\left({y}\right)=\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{2}}=\frac{−\sqrt{\mathrm{3}}}{\mathrm{3}}\Leftrightarrow{y}=−\frac{\pi}{\mathrm{6}}+{k}\pi \\ $$$${or}\:\:\:{tan}\left({y}\right)=\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}}\Leftrightarrow{y}=\frac{\pi}{\mathrm{3}}+{k}\pi \\ $$$$ \\ $$
Commented by tawa tawa last updated on 17/Sep/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$