Cho tam giác ABC. Khi đó:

Đáp án: a) Sai. b) Đúng. c) Đúng. d) Đúng.

a) Sai. \(\sin B \cdot \cos C + \sin C \cdot \cos B\)

= sin(B + C) = sin(180° − A) = sinA.

b) Đúng. cos²A + cos²B + cos²C

\( = \frac{{1 + \cos 2A}}{2} + \frac{{1 + \cos 2B}}{2} + {\cos ^2}C = 1 + \frac{{\cos 2A + \cos 2B}}{2} + {\cos ^2}C\)

\( = 1 + \frac{1}{2} \cdot 2\left[ {\cos \left( {A + B} \right) \cdot \cos \left( {A - B} \right)} \right] + {\cos ^2}C = 1 + \cos \left( {\pi - C} \right) \cdot \cos \left( {A - B} \right) + {\cos ^2}C\)

\( = 1 - \cos C \cdot \cos \left( {A - B} \right) + {\cos ^2}C = 1 - \cos C \cdot \left[ {\cos \left( {A - B} \right) - \cos C} \right]\)

\( = 1 - \cos C\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right] = 1 - 2\cos A \cdot \cos B \cdot \cos C\).

c) Đúng. \[\tan A + \tan B + \tan C\]

\( = \tan \left( {A + B} \right) \cdot \left( {1 - \tan A \cdot \tan B} \right) + \tan C = - \tan C \cdot \left( {1 - \tan A \cdot \tan B} \right) + \tan C\)

\( = - \tan C + \tan A \cdot \tan B \cdot \tan C + \tan C = \tan A \cdot \tan B \cdot \tan C\).

d) Đúng. \(\tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1\)

Ta có \(\tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2} \cdot \tan \frac{B}{2}}} \Leftrightarrow \frac{1}{{\tan \frac{C}{2}}} = \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2} \cdot \tan \frac{B}{2}}}\)

\( \Leftrightarrow \tan \frac{C}{2} \cdot \left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right) = 1 - \tan \frac{A}{2} \cdot \tan \frac{B}{2}\)

\( \Leftrightarrow \tan \frac{C}{2} \cdot \tan \frac{A}{2} + \tan \frac{C}{2} \cdot \tan \frac{B}{2} + \tan \frac{A}{2} \cdot \tan \frac{B}{2} = 1\).