Biết \(\sin \alpha + \cos \alpha = \dfrac{1}{5}\) và \(0 \le x \le \pi \) . Khi đó \(\tan \alpha \) bằng

* Hướng dẫn giải

Ta có:

\(1 = {\sin ^2}\alpha  + {\cos ^2}\alpha  \)

\(\;\;\;= {\sin ^2}\alpha  + {\left( {\dfrac{1}{5} - \sin \alpha } \right)^2} \)

\(\;\;\;= 2{\sin ^2}\alpha  - \dfrac{2}{5}\sin \alpha  + \dfrac{1}{{25}}\)

\( \Leftrightarrow 2{\sin ^2}\alpha  - \dfrac{2}{5}\sin \alpha  - \dfrac{{24}}{{25}} = 0 \)

\(\Leftrightarrow \left[ \begin{array}{l}\sin \alpha  = \dfrac{4}{5}\\\sin \alpha  =  - \dfrac{3}{5}\end{array} \right.\)

Do \(0 \le x \le \pi \) nên \(\sin \alpha  \ge 0\). Chọn \(\sin \alpha  = \dfrac{4}{5}\).

Suy ra \(\cos \alpha  = \dfrac{1}{5} - \dfrac{4}{5} =  - \dfrac{3}{5}\).

Vậy \(\tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} =  - \dfrac{4}{3}\).