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

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