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

Lời giải:

$\sin ^2a+\cos ^2a=1$

$\cos ^2a=1-\sin ^2a=1-(\frac{-5}{13})^2=\frac{144}{169}$

Vì $\pi < a< \frac{3\pi}{2}$ nên $\cos a< 0$

Do đó: $\cos a=-\sqrt{\frac{144}{169}}=\frac{-12}{13}$

$\sin 2a=2\sin a\cos a=2.\frac{-5}{13}.\frac{-12}{13}=\frac{120}{169}$

$\cos 2a=\cos ^2a-\sin ^2a=2\cos ^2a-1=2.\frac{144}{169}-1=\frac{119}{169}$

$\cos a=\cos ^2\frac{a}{2}-\sin ^2\frac{a}{2}$

$=1-2\sin ^2\frac{a}{2}$

$\Leftrightarrow \frac{-12}{13}=1-2\sin ^2\frac{a}{2}$

$\Rightarrow \sin ^2\frac{a}{2}=\frac{25}{26}$

Vì $\pi < a< \frac{3\pi}{2}$ nên $\sin \frac{a}{2}>0$

$\Rightarrow \sin \frac{a}{2}=\frac{5}{\sqrt{26}}$

\(sina+cosa=\sqrt{2}\left(\dfrac{\sqrt{2}}{2}sina+\dfrac{\sqrt{2}}{2}cosa\right)\)

\(=\left[{}\begin{matrix}\sqrt{2}\left(sina.cos\dfrac{\pi}{4}+cosa.sin\dfrac{\pi}{4}\right)\\\sqrt{2}\left(sina.sin\dfrac{\pi}{4}+cosa.cos\dfrac{\pi}{4}\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}\sqrt{2}sin\left(a+\dfrac{\pi}{4}\right)\\\sqrt{2}cos\left(a-\dfrac{\pi}{4}\right)\end{matrix}\right.\)

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

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

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

pi/2<a,b<pi

=>cos a<0; cos b<0; sin a>0; sin b>0

\(cosa=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5};sina=\sqrt{1-\left(-\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=-3/5:4/5=-3/4; tan b=12/13:(-5/13)=-12/5

\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana\cdot tanb}\)

\(=\dfrac{-\dfrac{3}{4}+\dfrac{-12}{5}}{1-\dfrac{-3}{4}\cdot\dfrac{-12}{5}}=\dfrac{63}{16}\)

sin(a-b)=sina*cosb-sinb*cosa

\(=\dfrac{3}{5}\cdot\dfrac{-5}{13}-\dfrac{-4}{5}\cdot\dfrac{12}{13}=\dfrac{-15+48}{65}=\dfrac{33}{65}\)

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

\(\Leftrightarrow2\cdot sin\left(\dfrac{a}{2}\right)\cdot cos\left(\dfrac{a}{2}\right)+2\cdot cos^2\left(\dfrac{a}{2}\right)-1-\dfrac{cos\left(\dfrac{a}{2}\right)}{sin\left(\dfrac{a}{2}\right)}=0\)

=>\(2\cdot cos\left(\dfrac{a}{2}\right)\left(sin\left(\dfrac{a}{2}\right)+cos\left(\dfrac{a}{2}\right)\right)=\dfrac{cos\left(\dfrac{a}{2}\right)+sin\left(\dfrac{a}{2}\right)}{sin\left(\dfrac{a}{2}\right)}\)

=>\(\left(cos\left(\dfrac{a}{2}\right)+sin\left(\dfrac{a}{2}\right)\right)\left(sin\left(a\right)-1\right)=0\)

=>cos(a/2)=-sin(a/2) hoặc sin a-1=0

=>cot(a/2)=-1 hoặc sina =1

=>a=-pi/2(loại) hoặc a=pi/2

\(tan\left(a+\dfrac{2013pi}{2}\right)=tan\left(a+\dfrac{pi}{2}\right)=tan\left(\dfrac{pi}{2}+\dfrac{pi}{2}\right)=tanpi=0\)