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Question Number 21241 by tawa tawa last updated on 17/Sep/17

Find y in 3rd quadrant tan(y − 30) = cot(y)

$$\mathrm{Find}\:\mathrm{y}\:\mathrm{in}\:\mathrm{3rd}\:\mathrm{quadrant} \\ $$$$\mathrm{tan}\left(\mathrm{y}\:−\:\mathrm{30}\right)\:=\:\mathrm{cot}\left(\mathrm{y}\right) \\ $$

Answered by sma3l2996 last updated on 17/Sep/17

tan(y−30)=cot(y) tan(y−30)=(1/(tan(y)))⇔((tan(y)−tan(30))/(1+tan(y)tan(30)))=(1/(tany)) with tan(30)=((√3)/3) tan(y)(tan(y)−tan(30))=1+tan(y)tan(30) tan^2 (y)−((√3)/3)tan(y)−((√3)/3)tan(y)=1 tan^2 (y)−2((√3)/3)tan(y)−1=0 Δ=(4/3)+4=((16)/3)=(((4(√3))/3))^2 tan(y)=((((2(√3))/3)−((4(√3))/3))/2)=((−(√3))/3)⇔y=−(π/6)+kπ or tan(y)=((((2(√3))/3)+((4(√3))/3))/2)=(√3)⇔y=(π/3)+kπ

$${tan}\left({y}−\mathrm{30}\right)={cot}\left({y}\right) \\ $$$${tan}\left({y}−\mathrm{30}\right)=\frac{\mathrm{1}}{{tan}\left({y}\right)}\Leftrightarrow\frac{{tan}\left({y}\right)−{tan}\left(\mathrm{30}\right)}{\mathrm{1}+{tan}\left({y}\right){tan}\left(\mathrm{30}\right)}=\frac{\mathrm{1}}{{tany}} \\ $$$${with}\:\:{tan}\left(\mathrm{30}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${tan}\left({y}\right)\left({tan}\left({y}\right)−{tan}\left(\mathrm{30}\right)\right)=\mathrm{1}+{tan}\left({y}\right){tan}\left(\mathrm{30}\right) \\ $$$${tan}^{\mathrm{2}} \left({y}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{tan}\left({y}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{tan}\left({y}\right)=\mathrm{1} \\ $$$${tan}^{\mathrm{2}} \left({y}\right)−\mathrm{2}\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{tan}\left({y}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{4}=\frac{\mathrm{16}}{\mathrm{3}}=\left(\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${tan}\left({y}\right)=\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{2}}=\frac{−\sqrt{\mathrm{3}}}{\mathrm{3}}\Leftrightarrow{y}=−\frac{\pi}{\mathrm{6}}+{k}\pi \\ $$$${or}\:\:\:{tan}\left({y}\right)=\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}}\Leftrightarrow{y}=\frac{\pi}{\mathrm{3}}+{k}\pi \\ $$$$ \\ $$

Commented by tawa tawa last updated on 17/Sep/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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