Question Number 111939 by Khanacademy last updated on 05/Sep/20
Answered by bemath last updated on 05/Sep/20
$${consider}\:\mathrm{tan}\:{x}+\mathrm{tan}\:{y}\:=\:\mathrm{4} \\ $$$$\Rightarrow\:\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\:+\:\frac{\mathrm{sin}\:{y}}{\mathrm{cos}\:{y}}\:=\:\mathrm{4} \\ $$$$\Rightarrow\:\frac{\mathrm{sin}\:\left({x}+{y}\right)}{\mathrm{cos}\:{x}.\mathrm{cos}\:{y}}\:=\:\mathrm{4}\:\Rightarrow\:\mathrm{sin}\:\left({x}+{y}\right)=\mathrm{4}×\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\left({x}+{y}\right)\:=\:\mathrm{6}\:.\:{If}\:{x},{y}\in\mathbb{R}\:,\:{it}\:{is}\: \\ $$$${impossible} \\ $$
Commented by aleks041103 last updated on 05/Sep/20
$${Since}\:{sin}\left({x}+{y}\right)=\mathrm{6},\:{then}: \\ $$$${cos}\left({x}+{y}\right)=\pm\sqrt{\mathrm{1}−\mathrm{6}^{\mathrm{2}} }=\pm\sqrt{−\mathrm{35}}=\pm{i}\sqrt{\mathrm{35}} \\ $$$${Then}: \\ $$$${tan}\left({x}+{y}\right)=\frac{{sin}\left({x}+{y}\right)}{{cos}\left({x}+{y}\right)}=\frac{\mathrm{6}}{\pm{i}\sqrt{\mathrm{35}}}=\mp{i}\sqrt{\frac{\mathrm{36}}{\mathrm{35}}} \\ $$$${tan}\left({x}+{y}\right)=\pm{i}\sqrt{\frac{\mathrm{36}}{\mathrm{35}}} \\ $$
Answered by $@y@m last updated on 05/Sep/20
$$\mathrm{0}\leqslant\mathrm{cos}\:{x}\leqslant\mathrm{1} \\ $$$$\mathrm{0}\leqslant\mathrm{cos}\:{y}\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant\mathrm{cos}\:{x}.\mathrm{cos}\:{y}\leqslant\mathrm{1} \\ $$$${So},\:{Please}\:{check}\:{the}\:{question}. \\ $$