Rút gọn biểu thức $B = \tan \alpha \left( {\dfrac{{1 + {{\cos }^2}\alpha }}{{\sin \alpha }} - \sin \alpha } \right)$ được:

$\sin \alpha  = \dfrac{{\sqrt 7 }}{4}$;${\rm{cos2}}\alpha  = \dfrac{1}{8}$

$\sin \alpha  =  - \dfrac{{\sqrt 7 }}{4}$;${\rm{cos2}}\alpha  = \dfrac{1}{8}$

$\sin \alpha  = \dfrac{{\sqrt 5 }}{4}$;${\rm{cos2}}\alpha  = \dfrac{{\sqrt {11} }}{4}$

$\sin \alpha  = \dfrac{{\sqrt 7 }}{4}$;${\rm{cos2}}\alpha  =  - \dfrac{1}{8}$

$- \dfrac{{3\sqrt 7 }}{8}$

$- \dfrac{{\sqrt 7 }}{8}$

$\dfrac{{3\sqrt 7 }}{8}$

$\dfrac{{\sqrt 7 }}{8}$

$\dfrac{{1 + m}}{2}$

$\dfrac{{1 - m}}{2}$

$\dfrac{{1 - m}}{5}$

$m$

$\cot \alpha$

$\tan \dfrac{\alpha }{2}$

$\sin 2\alpha$

$\tan \alpha$

${\sin ^2}\left( {\alpha  + \beta } \right)$

${\cos ^2}\left( {\alpha  + \beta } \right)$

${\tan ^2}\left( {\alpha  + \beta } \right)$

$\sin \left( {\alpha  + \beta } \right)$

$\sin \alpha$

$3$

$1$

$\cos \alpha$

$\dfrac{9}{7}$

$1$

$\dfrac{1}{2}$

$\dfrac{3}{5}$