Question Number 116009 by Khalmohmmad last updated on 30/Sep/20
$$\begin{cases}{\mathrm{tan}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{y}}\\{\mathrm{tan}\:\left(\mathrm{a}−\mathrm{b}\right)=\mathrm{x}}\end{cases}\:\:\:\mathrm{tan2a}=? \\ $$
Answered by bemath last updated on 30/Sep/20
$$\:\begin{cases}{{a}+{b}={r}}\\{{a}−{b}={s}}\end{cases}\Rightarrow\mathrm{2}{a}\:=\:{r}+{s} \\ $$$$\mathrm{tan}\:\left(\mathrm{2}{a}\right)\:=\:\frac{\mathrm{tan}\:{r}+\mathrm{tan}\:{s}}{\mathrm{1}−\mathrm{tan}\:{r}.\:\mathrm{tan}\:{s}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{x}+{y}}{\mathrm{1}−{xy}} \\ $$
Answered by bobhans last updated on 30/Sep/20
$$\mathrm{2}{a}\:=\:\left({a}+{b}\right)+\left({a}−{b}\right) \\ $$$$\mathrm{tan}\:\left(\mathrm{2}{a}\right)\:=\:\frac{\mathrm{tan}\:\left({a}+{b}\right)+\mathrm{tan}\:\left({a}−{b}\right)}{\mathrm{1}−\mathrm{tan}\:\left({a}+{b}\right).\mathrm{tan}\:\left({a}−{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{y}+{x}}{\mathrm{1}−{yx}} \\ $$