Cho \( \sin \alpha = - {4 \over 5}, \, \, \left( { \pi < \alpha < {{3 \pi } \over 2}} \right) \, \). Tính \( \cot \left( { \alpha - { \pi \over 4}} \right) \).
Giải chi tiết:

Ta có:
\(1 + {\cot ^2}\alpha = {1 \over {{{\sin }^2}\alpha }} \Leftrightarrow 1 + {\cot ^2}\alpha = {1 \over {{{\left( { - {4 \over 5}} \right)}^2}}} \Leftrightarrow {\cot ^2}\alpha = {9 \over {16}} \Leftrightarrow \left[ \matrix{ \cot \alpha = {3 \over 4} \hfill \cr \cot \alpha = - {3 \over 4} \hfill \cr} \right.\)
Vì \(\pi < \alpha < {{3\pi } \over 2} \Rightarrow \left\{ \matrix{ \sin \alpha < 0 \hfill \cr \cos \alpha < 0 \hfill \cr} \right. \Rightarrow \cot \alpha > 0 \Rightarrow \cot \alpha = {3 \over 4}\)
\(\cot \left( {\alpha - {\pi \over 4}} \right) = {{\tan \alpha \tan {\pi \over 4} + 1} \over {\tan \alpha - \tan {\pi \over 4}}} = {{{4 \over 3}.1 + 1} \over {{4 \over 3} - 1}} = 7\)
Chọn: B.