Cho \(\sin a + \cos a = \frac{7}{5}\). Tính tan a.

• Với cos a = 0.

Thứ lại vào \(\sin a + \cos a = \frac{7}{5}\) ta suy ra \(\sin a = \frac{7}{5}\) (không thỏa mãn)

• Với cos a ≠ 0.

\(\sin a + \cos a = \frac{7}{5}\) (1)

\( \Leftrightarrow {\left( {\sin a + \cos a} \right)^2} = \frac{{49}}{{25}}\)

\( \Leftrightarrow {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = \frac{{49}}{{25}}\)

\( \Leftrightarrow 1 + 2\sin a\cos a = \frac{{49}}{{25}}\)

\( \Leftrightarrow \sin a\cos a = \frac{{12}}{{25}}\)

\({\left( {\sin a - \cos a} \right)^2} = {\sin ^2}a + {\cos ^2}a - 2\sin a\cos a\)

\( = 1 - 2\,.\,\frac{{12}}{{25}} = \frac{1}{{25}}\)

\( \Rightarrow \sin a - \cos a = \pm \frac{1}{5}\) (2)

Từ (1) và (2) ta suy ra được:

\[\left[ \begin{array}{l}\left\{ \begin{array}{l}\sin a + \cos a = \frac{7}{5}\\\sin a - \cos a = \frac{1}{5}\end{array} \right.\\\left\{ \begin{array}{l}\sin a + \cos a = \frac{7}{5}\\\sin a - \cos a = - \frac{1}{5}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}\sin a = \frac{4}{5}\\\cos a = \frac{3}{5}\end{array} \right.\\\left\{ \begin{array}{l}\sin a = \frac{3}{5}\\\cos a = \frac{4}{5}\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\tan a = \frac{4}{3}\\\tan a = \frac{3}{4}\end{array} \right.\]