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: \(cos\left(a-\dfrac{pi}{6}\right)\)

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

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

y = \(\dfrac{sin^2x}{cosx\left(sinx-cosx\right)}+\dfrac{1}{4}\)

y = \(\dfrac{sin^2x}{sinx.cosx-cos^2x}+\dfrac{1}{4}=\dfrac{\dfrac{sin^2x}{cos^2x}}{\dfrac{sinx.cosx}{cos^2x}-1}+\dfrac{1}{4}\)

y = \(\dfrac{tan^2x}{tanx-1}+\dfrac{1}{4}\)

y = \(\dfrac{4tan^2x+tanx-1}{4tanx-4}\). Đặt t =  tanx. Do x ∈ \(\left(\dfrac{\pi}{4};\dfrac{\pi}{2}\right)\) nên t ∈ (1 ; +\(\infty\))\

Ta đươc hàm số f(t) = \(\dfrac{4t^2+t-1}{4t-4}\)

⇒ ymin = \(\dfrac{17}{4}\) khi t = 2. hay x = arctan(2) + kπ 

do a ∈ \(\left(0;\dfrac{\pi}{2}\right)\)⇒ \(\left\{{}\begin{matrix}sinx>0\\cosx>0\end{matrix}\right.\)

Mà tanx = 3 ⇒ \(\dfrac{sinx}{cosx}=3\Leftrightarrow\dfrac{sin^2x}{cos^2x}=9\Rightarrow10sin^2x=9\)

⇒ sinx = \(\dfrac{3}{\sqrt{10}}\)

⇒ sin (x + π) = -sinx = -\(\dfrac{3}{\sqrt{10}}\)

3/4pi<a<pi

=>sin a>0; cosa<0

sin2a=-4/5

=>2*sina*cosa=-4/5

=>sina*cosa=-2/5

(sina-cosa)^2=sin^2a+cos^2a-2*sina*cosa=1+4/5=9/5

=>sin a-cosa=3/căn 5

Chọn C

Áp dụng công thức tích phân từng phần ta có:

2tan a-cot a=1

=>2tana-1/tan a=1

=>\(\dfrac{2tan^2a-1}{tana}=1\)

=>2tan^2a-tana-1=0

=>(tan a-1)(2tana+1)=0

=>tan a=-1/2 hoặc tan a=1

\(P=\dfrac{tan\left(-a\right)+2\cdot cota}{3\cdot tan\left(\dfrac{pi}{2}+a\right)}=\dfrac{-tana+2\cdot cota}{-3\cdot cota}\)

TH1: tan a=-1/2

\(P=\dfrac{\dfrac{1}{2}+2\cdot\left(-2\right)}{-3\cdot\left(-2\right)}=-\dfrac{7}{2}:6=-\dfrac{7}{12}\)

TH2: tan a=1

=>cot a=1

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

Ta có :

\(2tan\alpha-cot\alpha=1\)

\(\Leftrightarrow2tan\alpha-\dfrac{1}{tan\alpha}=1\)

\(\Leftrightarrow2tan\alpha-\dfrac{1}{tan\alpha}-1=0\)

\(\Leftrightarrow\dfrac{2tan^2\alpha-tan\alpha-1}{tan\alpha}=0\left(tan\alpha\ne0\right)\)

\(\Leftrightarrow2tan^2\alpha-tan\alpha-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tan\alpha=1\\tan\alpha=-\dfrac{1}{2}\end{matrix}\right.\)

\(P=\dfrac{tan\left(8\pi-\alpha\right)+2cot\left(\pi+\alpha\right)}{3tan\left(\dfrac{3\pi}{2}+\alpha\right)}\)

\(\Leftrightarrow P=\dfrac{tan\left(4.2\pi-\alpha\right)+2cot\alpha}{3tan\left(2\pi-\dfrac{\pi}{2}+\alpha\right)}\)

\(\Leftrightarrow P=\dfrac{tan\left(-\alpha\right)+2cot\alpha}{3tan\left[-\left(\dfrac{\pi}{2}-\alpha\right)\right]}\)

\(\Leftrightarrow P=\dfrac{-tan\alpha+2cot\alpha}{-3tan\left(\dfrac{\pi}{2}-\alpha\right)}\)

\(\Leftrightarrow P=\dfrac{-tan\alpha+2cot\alpha}{-3cot\alpha}\)

- Với \(tan\alpha=1\Rightarrow cot\alpha=1\)

\(\Leftrightarrow P=\dfrac{-1+2.1}{-3.1}=-\dfrac{1}{3}\)

- Với \(tan\alpha=-\dfrac{1}{2}\Rightarrow cot\alpha=-2\)

\(\Leftrightarrow P=\dfrac{\dfrac{1}{2}+2.\left(-2\right)}{-3.\left(-2\right)}=\dfrac{-\dfrac{7}{2}}{6}=-\dfrac{7}{12}\)

`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`