Bài 11 trang 161 SGK Đại số 10

LG a

\(\tan A + \tan B  +  \tan C \)\(= \tan A\tan B\tan C, \) \(\left( {\widehat A,\;\widehat B,\;\widehat C \ne \frac{\pi }{2}} \right).\)

Lời giải chi tiết:

\(\begin{array}{l}
A + B + C = {180^0}\\
\Rightarrow A = {180^0} - \left( {B + C} \right)\\
\Rightarrow \tan A = \tan \left[ {{{180}^0} - \left( {B + C} \right)} \right]\\
= \tan \left[ { - \left( {B + C} \right)} \right] = - \tan \left( {B + C} \right)\\
= - \dfrac{{\tan B + \tan C}}{{1 - \tan B\tan C}}\\
= \dfrac{{\tan B + \tan C}}{{\tan B\tan C - 1}}\\
\Rightarrow \tan A = \dfrac{{\tan B + \tan C}}{{\tan B\tan C - 1}}\\
\Rightarrow \tan A\left( {\tan B\tan C - 1} \right) = \tan B + \tan C\\
\Leftrightarrow \tan A\tan B\tan C - \tan A = \tan B + \tan C\\
\Leftrightarrow \tan A\tan B\tan C = \tan A + \tan B + \tan C\\
\Rightarrow dpcm
\end{array}\)

Cách khác:

Vì A, B, C là ba góc của tam giác nên ta có : A + B + C = π.

⇒ C = π - (A + B); A + B = π - C

Ta có: tan A + tan B + tan C = (tan A + tan B) + tan C

= tan (A + B). (1 – tan A.tan B) + tan C

= tan (π – C).(1 – tan A. tan B) + tan C

= -tan C.(1 – tan A. tan B) + tan C

= -tan C + tan A. tan B. tan C + tan C

= tan A. tan B. tan C.

LG b

\(\sin 2A + \sin 2B + \sin 2C \)\(= 4\sin A\sin B\sin C\)

Lời giải chi tiết:

\(\begin{array}{l}
\sin 2A + \sin 2B + \sin 2C\\
= 2\sin \dfrac{{2A + 2B}}{2}\cos \dfrac{{2A - 2B}}{2} + \sin 2C\\
= 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + 2\sin C\cos C\\
= 2\sin \left( {{{180}^0} - C} \right)\cos \left( {A - B} \right) + 2\sin C\cos C\\
= 2\sin C\cos \left( {A - B} \right) + 2\sin C\cos C\\
= 2\sin C\left[ {\cos \left( {A - B} \right) + \cos C} \right]\\
= 2\sin C\left[ {\cos \left( {A - B} \right) + \cos \left( {{{180}^0} - \left( {A + B} \right)} \right)} \right]\\
= 2\sin C\left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]\\
= 2\sin C.\left[ { - 2\sin \dfrac{{A - B + A + B}}{2}\sin \dfrac{{A - B - A - B}}{2}} \right]\\
= - 4\sin C\sin A\sin \left( { - B} \right)\\
= - 4\sin A\sin C\left( { - \sin B} \right)\\
= 4\sin A\sin B\sin C
\end{array}\)

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