Question Number 166170 by mathls last updated on 14/Feb/22
$${tan}\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{17}}\:\:\:,\:\:\:{tan}\left({a}−{b}\right)=\frac{\mathrm{11}}{\mathrm{13}} \\ $$$${tan}\mathrm{2}{a}=?\:\:\:\:\:\:{tan}\mathrm{2}{b}=? \\ $$
Answered by cortano1 last updated on 14/Feb/22
$$\Rightarrow\mathrm{2}{a}\:=\:\left({a}+{b}\right)+\left({a}−{b}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{2}{a}\:=\frac{\frac{\mathrm{1}}{\mathrm{17}}+\frac{\mathrm{11}}{\mathrm{13}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{17}}.\frac{\mathrm{11}}{\mathrm{13}}}\:=\:\frac{\mathrm{20}}{\mathrm{21}} \\ $$
Commented by mathls last updated on 14/Feb/22
$${thanks} \\ $$
Answered by cortano1 last updated on 14/Feb/22
$$\Rightarrow\mathrm{2}{b}=\left({a}+{b}\right)−\left({a}−{b}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{2}{b}=\frac{\frac{\mathrm{1}}{\mathrm{17}}−\frac{\mathrm{11}}{\mathrm{13}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{17}}.\frac{\mathrm{11}}{\mathrm{13}}}=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$