Cho tam giác ABC.

Khi đó:

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

a) Đúng. Do A, B, C là ba góc trong tam giác nên A + B + C = 180°.

Vậy ta có tan(2A + B + C) = tan(A + 180°) = tan A.

b) Sai. Ta có cosB∙cosC = sinB∙sinC

\( \Leftrightarrow \frac{1}{2}\left[ {\cos \left( {B - C} \right) + \cos \left( {B + C} \right)} \right] = \frac{1}{2}\left[ {\cos \left( {B - C} \right) - \cos \left( {B + C} \right)} \right]\)

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

\( \Leftrightarrow \frac{1}{2}\cos \left( {B + C} \right) + \frac{1}{2}\cos \left( {B + C} \right) = 0 \Leftrightarrow \cos \left( {B + C} \right) = 0\)

B + C = 90°.

Vậy tam giác ABC vuông tại A.

c) Đúng. \(\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90\)°

\( \Rightarrow \tan \left( {\frac{A}{2} + \frac{B}{2} - \frac{C}{2}} \right) = \cot \left[ {90^\circ - \left( {\frac{A}{2} + \frac{B}{2} - \frac{C}{2}} \right)} \right] = \cot \left[ {\left( {\frac{A}{2} + \frac{B}{2} + \frac{C}{2}} \right) - \left( {\frac{A}{2} + \frac{B}{2} - \frac{C}{2}} \right)} \right]\)

= cotC.

d) Sai. \(4\cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{1}{2} \cdot 4\left[ {\cos \left( {\frac{A}{2} + \frac{B}{2}} \right) + \cos \left( {\frac{A}{2} - \frac{B}{2}} \right)} \right] \cdot \sin \frac{C}{2}\)

Vì A + B + C = 180 nên \(\frac{{A + B}}{2} = 90^\circ - \frac{C}{2}\) do đó \(\cos \left( {\frac{{A + B}}{2}} \right) = \sin \frac{C}{2}\).

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

= \(2{\sin ^2}\frac{C}{2} + 2\cos \left( {\frac{A}{2} - \frac{B}{2}} \right) \cdot \cos \left( {\frac{A}{2} + \frac{B}{2}} \right)\)

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

= 2\({\sin ^2}\frac{C}{2} + \cos A + \cos B\) = 1 − \(\cos C\) + cosA + cosB.