a/ Cho $\tan a + \cot a =m$ .  Tính theo $m$    $tan^{3} a + cot^{3} a, tan^{2} a + cot^{2} a$, $\left| {tan a - cot a} \right|$

b/ Cho $tan a = 3$ tính $\frac{2\sin a + 3\cos a}{4\sin a - 5\cos a} $, $\frac{3\sin a - 2\cos a}{5sin^{3}a + 4cos^{3}a}$

bai2: cho tam giac ABC co AB= 5, AC= 4 va dien tich bang $3\sqrt{3}$. tinh BC
Bài 2.
$S\triangle ABC=\frac{1}{2}AB.AC.sinA$
$\Rightarrow sinA=\frac{6\sqrt3}{4.5}=\frac{3\sqrt3}{10}$
$\Rightarrow cos^2a=1-\frac{27}{100}=\frac{73}{100}$
$\Rightarrow cosa=\pm\frac{\sqrt{73}}{10}$
TH1.  $cosa=\frac{\sqrt{73}}{10}$
$\Rightarrow BC^2=AB^2+AC^2-2AB.ACcosA=41-4\sqrt{73}$
$\Rightarrow BC=\sqrt{41-4\sqrt{73}}$
TH2.  $cosa=-\frac{\sqrt{73}}{10}$
$\Rightarrow BC^2=41+4\sqrt{73}$
$\Rightarrow BC=\sqrt{41+4\sqrt{73}}$
a.
$m^3=(tana+cota)^3=tan^3a+cot^3a+3tana.cota(tana+cota)$
$\Rightarrow m^3=tan^3a+cot^3a+3m$
$\Rightarrow tan^3a+cot^3a=m^3-3m$

$m^2=(tana+cota)^2=tan^2a+cot^2a+2tana.cota$
$\Rightarrow tan^2a+cot^2a=m^2-2$

$(tana-cota)^2=tan^2a+cot^2a-2tana.cota=m^2-4$
$\Rightarrow |tana=cota|=\sqrt{m^2-4}$

b.
$\frac{2sina+3cosa}{4sina-5cosa}=\frac{cosa(2tana+3)}{cosa(4tana-5)}=\frac{2.3+3}{4.3-5}=\frac{9}{7}$
$\frac{3sina-2cosa}{5sin^3a+4cos^3a}$
$=\frac{cosa(3tana-2)}{cos^3a(5tan^3a+4)}=\frac{1}{cos^2a}.\frac{3.3-2}{5.3^3+4}=\frac{1}{cos^2a}.\frac{7}{139}$
Vì $tana=3\Rightarrow sina=3cosa\Rightarrow sin^2a=9cos^2a$
Mà  $sin^a+cos^2a=1\Rightarrow cos^2a=\frac{1}{10}$
$\Rightarrow \frac{3sina-2cosa}{5sin^3a+4cos^3a}=\frac{70}{139}$

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