\(\dfrac{1+cos2a-sin2a}{1+cos2a+sin2a}=\dfrac{2cos^2a-2sina.cosa}{2cos^2a+2sinacosa}\)

\(=\dfrac{2cosa\left(cosa-sina\right)}{2cosa\left(cosa+sina\right)}=\dfrac{cosa-sina}{cosa+sina}=\dfrac{\sqrt{2}sin\left(\dfrac{\pi}{4}-a\right)}{\sqrt{2}cos\left(\dfrac{\pi}{4}-a\right)}=tan\left(\dfrac{\pi}{4}-a\right)\)

\(\dfrac{1+cos2a-cosa}{sin2a-sina}=\dfrac{2cos^2a-cosa}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)

a:

2: pi/2<a<pi

=>sin a>0 và cosa<0

tan a=-2

1+tan^2a=1/cos^2a=1+4=5

=>cos^2a=1/5

=>\(cosa=-\dfrac{1}{\sqrt{5}}\)

\(sina=\sqrt{1-\dfrac{1}{5}}=\dfrac{2}{\sqrt{5}}\)

cot a=1/tan a=-1/2

3: pi<a<3/2pi

=>cosa<0; sin a<0

1+cot^2a=1/sin^2a

=>1/sin^2a=1+9=10

=>sin^2a=1/10

=>\(sina=-\dfrac{1}{\sqrt{10}}\)

\(cosa=-\dfrac{3}{\sqrt{10}}\)

tan a=1:cota=1/3

b;

tan x=-2

=>sin x=-2*cosx

\(A=\dfrac{2\cdot sinx+cosx}{cosx-3sinx}\)

\(=\dfrac{-4cosx+cosx}{cosx+6cosx}=\dfrac{-3}{7}\)

2: tan x=-2 

=>sin x=-2*cosx

\(B=\dfrac{-4cosx+3cosx}{-6cosx-2cosx}=\dfrac{1}{8}\)

Có \(a\) thuộc góc phần tư thứ III -> sin\(a\) < 0

+) sin\(a\)=-\(\sqrt{1-cos^2a}\)=-\(\sqrt{1-\left(\dfrac{-12}{13}\right)^2}\)=\(\dfrac{-5}{13}\)

\(cos2a=cos^2a-sin^2a\)=\(\left(\dfrac{-12}{13}\right)^2-\left(\dfrac{-5}{13}\right)^2=\dfrac{119}{169}\)

\(5sin2a-6cosa=0\)

\(\Leftrightarrow sin2a=\dfrac{6}{5}cosa\)

\(\Leftrightarrow2\cdot sina\cdot cosa=\dfrac{6}{5}\cdot cosa\)

\(\Leftrightarrow cosa\left(2sina-\dfrac{6}{5}\right)=0\)

=>cosa=0 hoặc sina=3/5

hay \(a=\dfrac{\Pi}{2}+k\Pi\) hoặc \(\left[{}\begin{matrix}a=arcsin\left(\dfrac{3}{5}\right)+k2\Pi\\a=\Pi-arcsin\left(\dfrac{3}{5}\right)+k2\Pi\end{matrix}\right.\)

mà 0<a<pi/2

nên \(a=arcsin\left(\dfrac{3}{5}\right)\)

\(A=sina+sina+cota=2\cdot sina+cota\)

\(=\dfrac{38}{15}\)

bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)

\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)

Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(tan\alpha< 0,cot\alpha< 0;cos\alpha< 0\).
Vì vậy: \(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{7}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{4}:\dfrac{-\sqrt{7}}{4}=\dfrac{-3}{\sqrt{7}}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-\sqrt{7}}{3}\).
\(A=\dfrac{2tan\alpha-3cot\alpha}{cos\alpha+tan\alpha}\)\(=\dfrac{2.\dfrac{-3}{\sqrt{7}}-3.\dfrac{-\sqrt{7}}{3}}{\dfrac{-\sqrt{7}}{4}+\dfrac{-3}{\sqrt{7}}}\)
\(=\dfrac{\dfrac{-6}{\sqrt{7}}+\sqrt{7}}{\dfrac{-7-12}{4\sqrt{7}}}\)\(=\dfrac{\dfrac{-6+7}{\sqrt{7}}.4\sqrt{7}}{-19}\)\(=\dfrac{\dfrac{1}{\sqrt{7}}.4\sqrt{7}}{-19}=-\dfrac{4}{19}\).

b) \(\dfrac{cos^2\alpha+cot^2\alpha}{tan\alpha-cot\alpha}=\dfrac{\left(-\dfrac{\sqrt{7}}{4}\right)^2+\left(\dfrac{-\sqrt{7}}{3}\right)^2}{\dfrac{-3}{\sqrt{7}}+\dfrac{\sqrt{7}}{3}}\)
\(=\dfrac{\dfrac{7}{16}+\dfrac{7}{9}}{\dfrac{-9+7}{3\sqrt{7}}}=\dfrac{\dfrac{175}{144}}{\dfrac{-2}{3\sqrt{7}}}=\dfrac{-175}{96\sqrt{7}}\).