Cho tam giác \(ABC\). Tính giá trị của biểu thức:\...

D

- Hướng dẫn giải

Phương pháp giải:

Áp dụng các công thức sau: \(\left\{ \begin{array}{l}{\sin ^2}\alpha  + c{\rm{o}}{{\rm{s}}^2}\alpha  = 1\\\tan \alpha .\cot \alpha  = 1 \Rightarrow \tan \alpha .\frac{{\cos \alpha }}{{\sin \alpha }} = 1\\\sin \left( {{{90}^0} - \alpha } \right) = c{\rm{os}}\alpha \\\cos \left( {{{180}^0} - \alpha } \right) =  - c{\rm{os}}\alpha \end{array} \right..\)

Giải chi tiết:

Ta có: \(\angle A + \angle B + \angle C = {180^o} \Rightarrow \angle A + \angle C = {180^o} - \angle B\)

Lại có: \(\sin \left( {\frac{{{{180}^o} - \angle B}}{2}} \right) = \cos \frac{{\angle B}}{2};\,\,\,\,\cos \left( {\frac{{{{180}^o} - \angle B}}{2}} \right) = \sin \frac{{\angle B}}{2}.\)

  \(\begin{array}{l} \Rightarrow S = \frac{{{{\sin }^3}\frac{{\angle B}}{2}}}{{\cos \left( {\frac{{\angle A + \angle C}}{2}} \right)}} + \frac{{{{\cos }^3}\frac{{\angle B}}{2}}}{{\sin \left( {\frac{{\angle A + \angle C}}{2}} \right)}} - \frac{{\cos \left( {\angle A + \angle C} \right)}}{{\sin \angle B}}.\tan \angle B\\\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\sin }^3}\frac{{\angle B}}{2}}}{{\cos \left( {\frac{{{{180}^o} - \angle B}}{2}} \right)}} + \frac{{{{\cos }^3}\frac{{\angle B}}{2}}}{{\sin \left( {\frac{{{{180}^o} - \angle B}}{2}} \right)}} - \frac{{\cos \left( {{{180}^o} - \angle B} \right)}}{{\sin \angle B}}.\tan \angle B\\\,\,\,\,\,\,\,\,\, = \frac{{{{\sin }^3}\frac{{\angle B}}{2}}}{{\sin \frac{{\angle B}}{2}}} + \frac{{{{\cos }^3}\frac{{\angle B}}{2}}}{{\cos \frac{{\angle B}}{2}}} - \frac{{ - \cos \angle B}}{{\sin \angle B}}.\tan \angle B\\\,\,\,\,\,\,\,\,\, = {\sin ^2}\frac{{\angle B}}{2} + {\cos ^2}\frac{{\angle B}}{2} + \cot \angle B.\tan \angle B = 1 + 1 = 2.\end{array}\)

Chọn  D