Cho tam giác \(ABC.\) Tính \(M = {\tan ^2}\left( {...

C

- Hướng dẫn giải

Phương pháp giải:

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

Giải chi tiết:

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

\(\begin{array}{l} \Rightarrow \sin \left( {\angle C - {{90}^o}} \right) = \sin \left[ {{{180}^o} - \left( {\angle A + \angle B} \right) - {{90}^o}} \right] = \sin \left[ {{{90}^o} - \left( {\angle A + \angle B} \right)} \right] = \cos \left( {\angle A + \angle B} \right).\\\,\,\,\,\,\,\cot \left( {\angle C - {{90}^o}} \right) = \cot \left[ {{{180}^o} - \left( {\angle A + \angle B} \right) - {{90}^o}} \right] = \cot \left[ {{{90}^o} - \left( {\angle A + \angle B} \right)} \right] = \tan \left( {\angle A + \angle B} \right).\end{array}\)

Khi đó:

\(\begin{array}{l}M = {\tan ^2}\left( {\angle A + \angle B} \right).{\sin ^2}\left( {\angle C - {{90}^0}} \right) + 1:\left[ {1 + {{\cot }^2}\left( {\angle C - {{90}^0}} \right)} \right]\\\,\,\,\,\,\,\, = \frac{{{{\sin }^2}\left( {\angle A + \angle B} \right)}}{{{{\cos }^2}\left( {\angle A + \angle B} \right)}}.{\cos ^2}\left( {\angle A + \angle B} \right) + 1:\left[ {1 + {{\tan }^2}\left( {\angle A + \angle B} \right)} \right]\\\,\,\,\,\,\,\, = {\sin ^2}\left( {\angle A + \angle B} \right) + {\cos ^2}\left( {\angle A + \angle B} \right) = 1.\end{array}\)

Chọn  C