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

Đáp án đúng : B

Chọn C.

Theo giả thiết ta có:

P = ( sina + sinb) ² + ( cosa + cosb) ²

= sin²a + 2.sina.sinb + sin²b + cos²a + 2cosa. cosb + cos²b

= 2 + 2( sina.sinb + cos a. cosb)

= 2 + 2.cos( a - b)   ( sử dụng công thức cộng)

Ta có: \(tan\alpha=3=\frac{sin\alpha}{cos\alpha}\Rightarrow sin\alpha=3cos\alpha\)

Suy ra: \(B=\frac{\left(sin\alpha-cos\alpha\right)\left(sin^2\alpha+cos^2\alpha+sin\alpha.cos\alpha\right)}{\left(sin\alpha+cos\alpha\right)\left(sin^2\alpha+cos^2\alpha-sin\alpha.cos\alpha\right)}\)

\(=\frac{2cos\alpha.\left(1+3cos^2\alpha\right)}{4cos\alpha.\left(1-3cos^2\alpha\right)}=\frac{1+3cos^2\alpha}{2.\left(1-3cos^2\alpha\right)}\)

khó quá chị ơi

tan.a ?

Có \(\sin^2a+\cos^2a=1\)\(\Leftrightarrow\sin^2a=1-\cos^2a=1-\left(\frac{1}{3}\right)^2=\frac{8}{9}\)

\(\Leftrightarrow\sin a=\frac{\sqrt{8}}{3}\)

Xét  \(B=\frac{\sin a-3\cos a}{\sin a+2\cos a}=\frac{\frac{\sqrt{8}}{3}-3\cdot\frac{1}{3}}{\frac{\sqrt{8}}{3}+2\cdot\frac{1}{3}}=\frac{7-5\sqrt{2}}{2}\)

Chất rắn không tan : Cu 

\(m_{Cu}=3.2\left(g\right)\)

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(0.2......0.2...........................0.2\)

\(m_{hh}=m_{Cu}+m_{Zn}=3.2+0.2\cdot65=16.2\left(g\right)\)

\(m_{H_2SO_4}=0.2\cdot98=19.6\left(g\right)\)

\(b=m_{dd_{H_2SO_4}}=\dfrac{19.6}{20\%}=98\left(g\right)\)

Chia cả tử và mẫu cho \(cosa\)

\(D=\dfrac{\dfrac{cosa}{cosa}+\dfrac{sina}{cosa}}{\dfrac{cosa}{cosa}-\dfrac{sina}{cosa}}=\dfrac{1+tana}{1-tana}=\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}=3\)