Question

Помогите, срочно ! решить пример если tg(a) = 9
$\frac{18cos(a) - 12sin(a) +9}{sin(a) - 9 cos(a) + 3}$

Answer 1

Ответ:

$tga=9\\\\1+tg^2a=\dfrac{1}{cos^2a}\ \ \to \ \ 1+81=\dfrac{1}{cos^2a}\ \ \to \ \ cos^2a=\dfrac{1}{82}\ \ \to \ \ cosa=\pm \dfrac{1}{\sqrt{82}}\\\\\\\dfrac{18\, cosa-12\, sina+9}{sina-9\, cosa+3}=\dfrac{cosa\cdot \Big(18-12\cdot tga+\dfrac{9}{cosa}\Big)}{cosa\cdot \Big(tga-9+\dfrac{3}{cosa}\Big)}=\Big[\ cosa\ne 0\ \Big]=\\\\\\=\dfrac{18-12\cdot 9+\dfrac{9}{cosa}}{9-9+\dfrac{3}{cosa}}=\dfrac{\dfrac{9-90\, cosa}{cosa}}{\dfrac{3}{cosa}}=\dfrac{9-90\, cosa}{3}=3-30\, cosa=$

$=3-30\cdot \Big(\pm \dfrac{1}{\sqrt{82}}\Big)=\left[\begin{array}{cl}3-\dfrac{30}{\sqrt{82}}=\dfrac{3\cdot (\sqrt{82}-10)}{\sqrt{82}}\approx -0,31\\3+\dfrac{30}{\sqrt{82}} =\dfrac{3\cdot (\sqrt{82}+10)}{\sqrt{82}}\approx 6,31\end{array}\right$

Answer 2

Разделим числитель и знаменатель на cosα, подставим tgα=9;получим

(18-12tgα+9/cosα)/(tgα-9+(3/cosα))=(18-12*9+9/cosα)/(9-9+(3/cosα))=

9*(2-12+1/cosαα)/(3/cosα)=3cosα*(-10+1/cosα)=-30cosα+3;

cosα=±√1/(1+tg²α)=±√1/(1+81)=±1/√82;

-30cosα+3=-30*(±1/√82)+3

-30*(1/√82)+3=3-3.31=-0.31

-30*(-1/√82)+3≈3.313=6.31