Cho góc alpha thỏa mãn sinα = 12/13 và pi/2 < alpha < pi. Tính P = tan(pi/4 − alpha).

Chọn B

Ta có: \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = 1 - {(\frac{{12}}{{13}})^2} = \frac{{25}}{{169}} \Rightarrow \cos \alpha  =  \pm \frac{5}{{13}}\)

Vì: \(\frac{\pi }{2} < \alpha  < \pi \)nên \(\cos \alpha  < 0\)\( \Rightarrow \cos \alpha  =  - \frac{5}{{13}}\)vậy \[\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{{12}}{{13}}}}{{ - \frac{5}{{13}}}} =  - \frac{{12}}{5}\]

Vậy \[P = \tan \left( {\frac{\pi }{4} - \alpha } \right) = \frac{{\tan \frac{\pi }{4} - \tan \alpha }}{{1 + \tan \frac{\pi }{4}.\tan \alpha }} = \frac{{1 + \frac{{12}}{5}}}{{1 - \frac{{12}}{5}}} =  - \frac{{17}}{7}\].