Môn Toán Lớp 10 Tam giác ABC có a + c = 2b. Tính P=tan$\frac{A}{2}$tan $\frac{C}{2}$
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Athena
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Đáp án:
$\tan\dfrac A2\tan\dfrac C2=\dfrac13\Leftrightarrow a + c = 2b$
Giải thích các bước giải:
Ta có:
$\quad a + c = 2b$
$\Leftrightarrow 2R\sin A + 2R\sin C = 4R\sin B$
$\Leftrightarrow \sin A + \sin C = 2\sin B$
$\Leftrightarrow 2\sin\dfrac{A + C}{2}\cdot\cos\dfrac{A – C}{2}= 2\sin(A + C)$
$\Leftrightarrow \sin\dfrac{A + C}{2}\cdot\cos\dfrac{A – C}{2} = 2\sin\dfrac{A+C}{2}\cdot\cos\dfrac{A+C}{2}$
$\Leftrightarrow \cos\dfrac{A – C}{2} = 2\cos\dfrac{A+C}{2}$
$\Leftrightarrow \cos\dfrac A2\cos\dfrac C2+ \sin\dfrac A2\sin\dfrac C2 = 2\cos\dfrac A2\cos\dfrac C2 – 2\sin\dfrac A2\sin\dfrac C2$
$\Leftrightarrow 3\sin\dfrac A2\sin\dfrac C2 = \cos\dfrac A2\cos\dfrac C2$
$\Leftrightarrow \dfrac{\sin\dfrac A2\sin\dfrac C2}{\cos\dfrac A2\cos\dfrac C2}=\dfrac13$
$\Leftrightarrow \tan\dfrac A2\tan\dfrac C2=\dfrac13$
Đáp án:$P=\dfrac{1}{3}$
Giải thích các bước giải:
Ta có :
$tan\dfrac{a}{2}.tan\dfrac{c}{2}\\=a+c=2b\\\to sina+sinc=2Sin B\\\to sin\dfrac{a+c}{2}sin\dfrac{a-c}{2}=2sin(a+c)\\\Leftrightarrow \cos\dfrac{a-c}{2} = 2\cos\dfrac{a+c}{2}\\\Leftrightarrow \cos\dfrac{a}{2}\cos\dfrac{c}{2}+ \sin\dfrac{a}{2}\sin\dfrac {c}{2} = 2\cos\dfrac{a}{2}\cos\dfrac {c}{2} – 2\sin\dfrac{a}{2}\sin\dfrac{c}{2}\\=\dfrac{\sin\dfrac{a}{2}\sin\dfrac{c}{2}}{\cos\dfrac{a}{2}\cos\dfrac{c}{2}}\\\toP=\dfrac{1}{3}$