Biết \(\tan\left(\alpha+\beta\right)=5;\tan\left(\alpha-\beta\right)=3\). Giá trị đúng của \(\tan2\alpha\) là
\(\dfrac{7}{4}\).\(-\dfrac{7}{4}\).\(\dfrac{4}{7}\).\(-\dfrac{4}{7}\).Hướng dẫn giải:\(\tan2\alpha=\tan\left[\left(\alpha+\beta\right)+\left(\alpha-\beta\right)\right]\)
\(=\dfrac{\tan\left(\alpha+\beta\right)+\tan\left(\alpha-\beta\right)}{1-\tan\left(\alpha+\beta\right).\tan\left(\alpha-\beta\right)}=\dfrac{5+3}{1-5.3}=\dfrac{8}{-14}=-\dfrac{4}{7}\)
