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 $

\(\dfrac{9}{7}\)

$1$

\(\dfrac{1}{2}\)

\(\dfrac{3}{5}\)