Đáp án: $\tan a=-1, \tan2b=\dfrac17$

Giải thích các bước giải:

Ta có:

$\begin{cases}\tan(a+b)=3\\\tan(a-b)=2\end{cases}$

$\to \begin{cases}\dfrac{\tan a+\tan b}{1-\tan a\tan b}=3\\\dfrac{\tan a-\tan b}{1+\tan a\tan b}=2\end{cases}$

Đặt $\tan a=x,\tan b=y$

$\to \begin{cases}\dfrac{x+y}{1-xy}=3\\\dfrac{x-y}{1+xy}=2\end{cases}$

$\to \begin{cases}\dfrac{x+y}3=1-xy\\ \dfrac{x-y}2=1+xy\end{cases}$

$\to \begin{cases}\dfrac{x+y}3+\dfrac{x-y}2=1-xy+1+xy\\ \dfrac{x-y}2=1+xy\end{cases}$

$\to \begin{cases}\dfrac{5x-y}{6}=2\\ \dfrac{x-y}2=1+xy\end{cases}$

$\to \begin{cases}5x-y=12\\ \dfrac{x-y}2=1+xy\end{cases}$

$\to \begin{cases}y=5x-12\\ \dfrac{x-(5x-12)}2=1+x(5x-12)\end{cases}$

$\to \begin{cases}y=5x-12\\ x=1\pm\sqrt2\end{cases}$

$\to (x,y)\in\{ (1+\sqrt2, -7+5\sqrt2), (1-\sqrt2, -7-5\sqrt2)\}$

$\to (\tan a,\tan b)\in\{ (1+\sqrt2, -7+5\sqrt2), (1-\sqrt2, -7-5\sqrt2)\}$

$\to \tan(2a, \tan2b)\in\{(-1, \dfrac17, ), (-1, \dfrac17)\}$