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

mà \(\sin x>0\)

nên \(\sin x=\dfrac{3}{5}\)

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

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

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 : \(tan\alpha=\dfrac{2}{\sqrt{21}},cot\alpha=\dfrac{\sqrt{21}}{2}\)

Ta có sin   α = 3 5 suy ra sin 2 α = 9 25 , mà  sin 2 α + cos 2 α = 1 , do đó:

cos 2 α = 1 - sin 2 α = 1 - 9 25 = 16 25 suy ra cos   α = 4 5

Do đó:

tan   α = sin α cos α = 3 5 : 4 5 = 3 5 . 5 4 = 3 4

c o t   α = cos α sin α = 4 5 : 3 5 = 4 5 . 5 3 = 4 3

Vậy cos   α = 4 5 ; tan   α = 3 4 ; c o t   α = 4 3

Đáp án cần chọn là: B

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

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

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

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

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

mà \(\cos x< 0\)

nên \(\cos x=-\dfrac{4}{5}\)

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

\(tanx=-2\sqrt{2}\)

=>\(cotx=\dfrac{1}{-2\sqrt{2}}=\dfrac{-\sqrt{2}}{4}\)

\(tanx=-2\sqrt{2}\)

=>\(\dfrac{sinx}{cosx}=-2\sqrt{2}\)

=>sin x và cosx khác dấu 

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

=>\(\dfrac{1}{cos^2x}=1+8=9\)

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

=>\(\left[{}\begin{matrix}cosx=\dfrac{1}{3}\\cosx=-\dfrac{1}{3}\end{matrix}\right.\)

TH1: \(cosx=\dfrac{1}{3}\)

\(tanx=-2\sqrt{2}\)

=>\(\dfrac{sinx}{cosx}=-2\sqrt{2}\)

=>\(sinx=-2\sqrt{2}\cdot cosx=-2\sqrt{2}\cdot\dfrac{1}{3}=-\dfrac{2\sqrt{2}}{3}\)

TH2: cosx=-1/3

=>\(sinx=-2\sqrt{2}\cdot cosx=-2\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{2\sqrt{2}}{3}\)

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

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

a: Sửa đề: sin x=4/5

cosx=-3/5; tan x=-4/3; cot x=-3/4

b: 270 độ<x<360 độ

=>cosx>0

=>cosx=1/2

tan x=căn 3; cot x=1/căn 3

\(\cos\alpha=\sqrt{1-\dfrac{1}{25}}=\dfrac{2\sqrt{6}}{5}\)

\(\tan\alpha=\dfrac{1}{5}:\dfrac{2\sqrt{6}}{5}=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)

\(\cot\alpha=1:\dfrac{1}{2\sqrt{6}}=2\sqrt{6}\)