luong giac giup minh voi! can gap

Question

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

score 0 · Answer 1

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.

$cosa=-\frac{\sqrt{73}}{10}$

$\Rightarrow BC^2=41+4\sqrt{73}$

$\Rightarrow BC=\sqrt{41+4\sqrt{73}}$

score 0 · Answer 2

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

$sin^a+cos^2a=1\Rightarrow cos^2a=\frac{1}{10}$

$\Rightarrow \frac{3sina-2cosa}{5sin^3a+4cos^3a}=\frac{70}{139}$

2 Đáp án