Câu 1.
Ta có \({\tan ^2}\alpha = \dfrac{1}{{{{\cos }^2}\alpha }} - 1 = \dfrac{{1681}}{{81}} - 1\)\(\; = \dfrac{{1600}}{{81}}\).
Mà \(\pi < \alpha < \dfrac{{3\pi }}{2}\) nên \(\tan \alpha < 0\). Suy ra \(\tan \alpha = \dfrac{{40}}{9}\).
Do đó
\(\tan \left( {\alpha - \dfrac{{3\pi }}{4}} \right) = \dfrac{{\tan \alpha - \tan \dfrac{{3\pi }}{4}}}{{1 + \tan \alpha \tan \dfrac{{3\pi }}{4}}} \)
\(= \dfrac{{\tan \alpha + 1}}{{1 - \tan \alpha }} = \dfrac{{\dfrac{{40}}{9} + 1}}{{1 - \dfrac{{40}}{9}}} = - \dfrac{{49}}{{31}}\).
Câu 2. Ta có
\(\tan 45^\circ = \tan \left( {20^\circ + 25^\circ } \right)\)\(\; = \dfrac{{\tan 20^\circ + \tan 25^\circ }}{{1 - \tan 20^\circ \tan 25^\circ }}\).
Suy ra: \(1 - \tan 20^\circ \tan 25^\circ \)\(\;= \tan 20^\circ + \tan 25^\circ \)
\( \Leftrightarrow 1 + \tan 20^\circ + \tan 25^\circ + \tan 20^\circ \tan 25^\circ = 2\)
Vậy \(\left( {1 + \tan 20^\circ } \right)\left( {1 + \tan 25^\circ } \right) = 2\).
