\(a,,0< x< \dfrac{\pi}{2}\\ \Rightarrow\left\{{}\begin{matrix}sinx>0\\cosx< 0\end{matrix}\right.\\ 1+tan^2x=\dfrac{1}{cos^2x}\\ \Rightarrow cos^2x=\dfrac{1}{4}\\ \Rightarrow cosx=-\dfrac{1}{2}\)

\(sin^2x+cos^2x=1\\ \Rightarrow sin^2x=1-\left(-\dfrac{1}{2}\right)^2\\ =\dfrac{3}{4}\\ \Rightarrow sinx=\dfrac{\sqrt{3}}{2}\)

\(tanx.cotx=1\\ \Rightarrow cotx=1:\sqrt{3}\\ =\dfrac{\sqrt{3}}{3}\)

\(b,\dfrac{3\pi}{2}< x< 2\pi\\ \Rightarrow\left\{{}\begin{matrix}sinx< 0\\cosx>0\end{matrix}\right.\)

\(tanx.cotx=1\\ \Rightarrow tanx=-1\)

\(1+cot^2x=\dfrac{1}{sin^2x}\\ \Rightarrow sin^2x=\dfrac{1}{2}\\ \Rightarrow sinx=-\dfrac{\sqrt{2}}{2}\\ cos^2x+sin^2x=1\\ \Rightarrow cos^2x=\dfrac{1}{2}\\ \Rightarrow cosx=\dfrac{\sqrt{2}}{2}\)

a: pi<x<3/2pi

=>sinx<0 và cosx<0

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+\dfrac{9}{4}=\dfrac{13}{4}\)

=>\(cos^2x=\dfrac{4}{13}\)

=>\(\left\{{}\begin{matrix}cosx=-\dfrac{2}{\sqrt{13}}\\sin^2x=\dfrac{9}{13}\end{matrix}\right.\)

mà sin x<0

nên \(sinx=-\dfrac{3}{\sqrt{13}}\)

\(cotx=1:\dfrac{3}{2}=\dfrac{2}{3}\)

b: 0<x<90 độ

=>sin x>0 và cosx>0

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+\dfrac{1}{3}=\dfrac{4}{3}\)

=>\(cos^2x=\dfrac{3}{4}\)

=>\(cosx=\dfrac{\sqrt{3}}{2}\)

=>\(sinx=\dfrac{1}{2}\)

cotx=1:căn 3/3=3/căn 3=căn 3

c: 3/2pi<x<2pi

=>sinx<0 và cosx>0

\(1+cot^2x=\dfrac{1}{sin^2x}\)

=>\(\dfrac{1}{sin^2x}=1+\dfrac{1}{3}=\dfrac{4}{3}\)

=>\(sin^2x=\dfrac{3}{4}\)

mà sin x<0

nên \(sinx=-\dfrac{\sqrt{3}}{2}\)

\(cos^2x=1-\dfrac{3}{4}=\dfrac{1}{4}\)

mà cosx>0

nên cosx=1/2

a.Ta có : \(x\in\left(\pi;\dfrac{3}{2}\pi\right)\Rightarrow cosx< 0\) 

\(cosx=-\sqrt{1-sin^2x}=-\sqrt{1-0,8^2}=-0,6\) 

\(tanx=\dfrac{4}{3};cotx=\dfrac{3}{4}\)

b. cos 2x = \(cos^2x-sin^2x=0,6^2-0,8^2=-0,28\)

\(P=2.cos2x=-0,56\)

\(Q=tan\left(2x+\dfrac{\pi}{3}\right)=\dfrac{tan2x+tan\dfrac{\pi}{3}}{1-tan2x.tan\dfrac{\pi}{3}}=\dfrac{tan2x+\sqrt{3}}{1-tan2x.\sqrt{3}}\)

tan 2x = \(\dfrac{2tanx}{1-tan^2x}=\dfrac{\dfrac{2.4}{3}}{1-\left(\dfrac{4}{3}\right)^2}=\dfrac{-24}{7}\) 

\(Q=\dfrac{-\dfrac{24}{7}+\sqrt{3}}{1+\dfrac{24}{7}.\sqrt{3}}\) \(=\dfrac{-24+7\sqrt{3}}{7+24\sqrt{3}}\) 

\(\sin^2x=\sqrt{1-\left(-\dfrac{4}{5}\right)^2}=\dfrac{9}{25}\)

mà \(\sin x>0\)

nên \(\sin x=\dfrac{3}{5}\)

=>\(\tan x=-\dfrac{3}{4}\)

\(\Leftrightarrow\cot x=-\dfrac{4}{3}\)

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

\(\dfrac{\pi}{2}< x< \pi\Rightarrow cosx< 0\)

\(\Rightarrow cosx=-\sqrt{1-sin^2x}=-\dfrac{20}{29}\)

\(tanx=\dfrac{sinx}{cosx}=-\dfrac{21}{20}\)

\(cotx=\dfrac{1}{tanx}=-\dfrac{20}{21}\)

Em 2k8 ms học nên k chắc

Vì 0 < \(\alpha< \dfrac{\pi}{2}\)  => sin \(\alpha>0\)

Cos \(\alpha=\dfrac{1}{3}\)  \(\Rightarrow sin\alpha=\sqrt{1-\dfrac{1}{9}}=\dfrac{2\sqrt{2}}{3}\)

tan \(\alpha=2\sqrt{2}\)  ; cot \(\alpha=\dfrac{1}{2\sqrt{2}}\)

b)\(P=cos2a-cos(\dfrac{\pi}{3}-a) \\=2cos^2a-1-cos\dfrac{\pi}{3}cosa-sin\dfrac{\pi}{3}sina \\=2.(\dfrac{-2}{5})^2-1-\dfrac{1}{2}.\dfrac{-2}{5}-\dfrac{\sqrt3}{2}.\dfrac{-\sqrt{21}}{5} \\=\dfrac{-24+15\sqrt7}{50}\)

a, Vì : \(\pi< a< \dfrac{3\pi}{2}\)  nên \(cos\alpha< 0\) mà \(cos^2\alpha=1-sin^2\alpha=1-\dfrac{4}{25}=\dfrac{21}{25},\)

do đó : \(cos\alpha=-\dfrac{\sqrt{21}}{5}\)

từ đó suy ra : \(tan\alpha=\dfrac{2}{\sqrt{21}},cot\alpha=\dfrac{\sqrt{21}}{2}\)

\(\cos^2x=\sqrt{1-\dfrac{9}{25}}=\dfrac{16}{25}\)

mà \(\cos x< 0\)

nên \(\cos x=-\dfrac{4}{5}\)

=>\(\tan x=-\dfrac{3}{4};\cot x=-\dfrac{4}{3}\)