a: -pi/2<a<0

=>sin a<0

=>sin a=-1/căn 5

tan a=-1/2

cot a=-2

b: pi/2<x<pi

=>cosx<0

=>cosx=-4/5

=>tan x=-3/4

cot x=-4/3

c: -pi<x<-pi/2

=>cosx<0 và sin x<0

1+tan^2x=1/cos^2x

=>1/cos^2x=1+16/25=41/25

=>cosx=-5/căn 41

sin x=-6/căn 41

cot x=5/4

g: 180 độ<x<270 độ

=>cosx <0

=>cosx=-4/5

tan x=3/4

cot x=4/3

a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cot\alpha>0;tan\alpha>0\).
Vì vậy: \(sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{15}}{4}:\dfrac{-1}{4}=\sqrt{15}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\sqrt{15}}\).

b) Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).

c) Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
Có \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{1}{tan^2\alpha+1}\)
Vì vậy: \(cos\alpha=\sqrt{\dfrac{1}{tan^2+1}}=\dfrac{\sqrt{58}}{3}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{3}{7}\).
\(sin\alpha=cos\alpha:cot\alpha=\dfrac{\sqrt{58}}{3}:\dfrac{3}{7}=\dfrac{7\sqrt{58}}{9}\).

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

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

Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha,cos\alpha< 0;tan\alpha,cot\alpha< 0\).
\(cos\left(\alpha-\dfrac{\pi}{2}\right)=cos\left(\dfrac{\pi}{2}-\alpha\right)=sin\alpha< 0\).
\(sin\left(\dfrac{\pi}{2}+\alpha\right)=cos\alpha< 0\).
\(tan\left(\dfrac{3\pi}{2}-\alpha\right)=tan\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)\)\(=tan\left(-\dfrac{\pi}{2}-\alpha\right)\)\(=-tan\left(\dfrac{\pi}{2}+\alpha\right)=cot\left(\alpha\right)>0\).
\(cot\left(\alpha+\pi\right)=cot\left(\alpha\right)>0\).

\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)

\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)

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

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

\(\dfrac{\pi}{2}< a< \pi\Rightarrow sina>0\)

\(\Rightarrow sina=\sqrt{1-cos^2a}=\dfrac{\sqrt{5}}{3}\)

\(K=2sina.cosa+2cos^2a-1=-\dfrac{1}{9}-\dfrac{4}{9}\sqrt{5}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{1}{4}\Rightarrow a-b=-3\)

b) Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
Vì vậy:
\(cos\alpha=\sqrt{1-0,6^2}=\dfrac{4}{5}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=0,6:\dfrac{4}{5}=0,75;cot\alpha=1:tan\alpha=\dfrac{4}{3}\).

Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(sin\alpha>0;tan\alpha< 0;cot\alpha< 0\).
\(sin\alpha=\sqrt{1-cos^2\alpha}=\dfrac{\sqrt{51}}{10}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{51}}{10}:\left(-0,7\right)=-\dfrac{\sqrt{51}}{7}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-7}{\sqrt{51}}\).

Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cos\alpha< 0;cot\alpha>2\).
\(cos\alpha=-\sqrt{\dfrac{1}{tan^2+1}}=-\dfrac{1}{\sqrt{5}}\);
\(cot\alpha=1:tan\alpha=\dfrac{1}{2}\);
\(sin\alpha=-\sqrt{1-cos^2\alpha}=-\dfrac{2}{\sqrt{5}}\).