Bài 3 trang 20

Đề bài

Cho $\tan \left(a+b\right)=3,\tan \left(a-b\right)=2$.

Tính: $\tan 2a,\tan 2b$

Lời giải chi tiết

Ta có:

$\begin{array}{c}2a=\left(a+b\right)+\left(a-b\right)\Rightarrow \tan 2a=\tan \left[\left(a+b\right)+\left(a-b\right)\right]\\ 2b=\left(a+b\right)-\left(a-b\right)\Rightarrow \tan 2b=\tan \left[\left(a+b\right)-\left(a-b\right)\right]\end{array}$

$\begin{array}{c}\tan \left[\left(a+b\right)+\left(a-b\right)\right]=\frac{\tan \left(a+b\right)+\tan \left(a-b\right)}{1-\tan \left(a+b\right).\tan \left(a-b\right)}=\frac{3+2}{1-3.2}=-1\\ \tan \left[\left(a+b\right)-\left(a-b\right)\right]=\frac{\tan \left(a+b\right)-\tan \left(a-b\right)}{1+\tan \left(a+b\right).\tan \left(a-b\right)}=\frac{3-2}{1+3.2}=\frac{1}{7}\end{array}$

Vậy $\tan 2a=-1,\tan 2b=\frac{1}{7}$