`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`

a: pi/2<a<pi

=>sin a>0

\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)

\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)

\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)

b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)

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

c: \(sin\left(a-\dfrac{pi}{3}\right)\)

\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)

\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)

d: \(cos\left(a-\dfrac{pi}{6}\right)\)

\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)

Bài 1 :

Ta có : a thuộc góc phần tư thứ II .

=> Cos a < 0

- Ta lại có : \(\left\{{}\begin{matrix}sina=\dfrac{1}{3}\\sin^2a+cos^2a=1\end{matrix}\right.\)

\(\Rightarrow cosa=\sqrt{1-\left(\dfrac{1}{3}\right)^2}=-\dfrac{2\sqrt{2}}{3}\)

Bài 2 :

Ta có : \(F=\dfrac{\cos x.\tan x}{\sin^2x-\cot x.\cos x}=\dfrac{\cos x.\dfrac{\sin x}{\cos x}}{\sin^2x-\dfrac{\cos x}{\sin x}.\cos x}\)

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

Chọn C.

Ta có :

P = sin(π + α).cos(π - α) = -sin α.(-cos α) = sin α.cos α.

Và  = cos α.(-sin α) = -sin α.cos α.

Do đó; P + Q = 0.

1: 

a: sin a=căn 3/2

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

cot a=1/tan a=1/căn 3

b: \(tana=2\)

=>cot a=1/tan a=1/2

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

=>\(\dfrac{1}{cos^2a}=5\)

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)

c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=5/13:12/13=5/12

cot a=1:5/12=12/5

a: VT=sin^2a(sin^2a+cos^2a)+cos^2a

=sin^2a+cos^2a

=1=VP

b: \(VT=\dfrac{sina+sina\cdot cosa+sina-sina\cdot cosa}{1-cos^2a}=\dfrac{2sina}{sin^2a}=\dfrac{2}{sina}=VP\)

c: \(VT=\dfrac{sin^2a+1+2cosa+cos^2a}{sina\left(1+cosa\right)}\)

\(=\dfrac{2\left(cosa+1\right)}{sina\left(1+cosa\right)}=\dfrac{2}{sina}=VP\)

a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)

\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)

b: Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)

\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)

=4