Đáp án:

 B1:

a) $\sin \alpha  = \dfrac{{\sqrt 7 }}{3};\tan \alpha  = \dfrac{{ - \sqrt 7 }}{{\sqrt 2 }};\cot \alpha  = \dfrac{{ - \sqrt 2 }}{{\sqrt 7 }}$

b) $\cos \alpha  = \dfrac{{ - 1}}{3};\sin \alpha  = \dfrac{{ - 2\sqrt 2 }}{3};\cot \alpha  = \dfrac{1}{{2\sqrt 2 }}$

c) $\cos \alpha  = \dfrac{{ - \sqrt 5 }}{3};\tan \alpha  = \dfrac{2}{{\sqrt 5 }};\cot \alpha  = \dfrac{{\sqrt 5 }}{2}$

d) $\sin \alpha  = \dfrac{{\sqrt {15} }}{4};\tan \alpha  =  - \sqrt {15} ;\cot \alpha  = \dfrac{{ - 1}}{{\sqrt {15} }}$

B2:

$\begin{array}{l}
a)A = {\tan ^2}\alpha \\
b)B = \cot 2a
\end{array}$

Giải thích các bước giải:

B1:

$a)\cos \alpha  = \dfrac{{ - \sqrt 2 }}{3}$

$\begin{array}{l}
Do:\dfrac{\pi }{2} < \alpha  < \pi  \Rightarrow \sin \alpha  > 0\\
 \Rightarrow \sin \alpha  = \sqrt {1 - {{\cos }^2}\alpha }  = \dfrac{{\sqrt 7 }}{3}\\
 \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{{ - \sqrt 7 }}{{\sqrt 2 }};\cot \alpha  = \dfrac{1}{{\tan \alpha }} = \dfrac{{ - \sqrt 2 }}{{\sqrt 7 }}
\end{array}$

Vậy $\sin \alpha  = \dfrac{{\sqrt 7 }}{3};\tan \alpha  = \dfrac{{ - \sqrt 7 }}{{\sqrt 2 }};\cot \alpha  = \dfrac{{ - \sqrt 2 }}{{\sqrt 7 }}$

$\begin{array}{l}
b)\tan \alpha  = 2\sqrt 2 \\
 \Rightarrow \cot \alpha  = \dfrac{1}{{2\sqrt 2 }}\\
Do:\pi  < \alpha  < \dfrac{{3\pi }}{2} \Rightarrow \cos \alpha  < 0\\
 \Rightarrow \cos \alpha  =  - \sqrt {\dfrac{1}{{1 + {{\tan }^2}\alpha }}}  =  - \dfrac{1}{3}\\
 \Rightarrow \sin \alpha  = \tan \alpha .\cos \alpha  = \dfrac{{ - 2\sqrt 2 }}{3}
\end{array}$

Vậy $\cos \alpha  = \dfrac{{ - 1}}{3};\sin \alpha  = \dfrac{{ - 2\sqrt 2 }}{3};\cot \alpha  = \dfrac{1}{{2\sqrt 2 }}$

$\begin{array}{l}
c)\sin \alpha  = \dfrac{{ - 2}}{3}\\
Do:\dfrac{{3\pi }}{2} < \alpha  < 2\pi  \Rightarrow \cos \alpha  > 0\\
 \Rightarrow \cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha }  = \dfrac{{ - \sqrt 5 }}{3}\\
 \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{2}{{\sqrt 5 }};\cot \alpha  = \dfrac{1}{{\tan \alpha }} = \dfrac{{\sqrt 5 }}{2}
\end{array}$

Vậy $\cos \alpha  = \dfrac{{ - \sqrt 5 }}{3};\tan \alpha  = \dfrac{2}{{\sqrt 5 }};\cot \alpha  = \dfrac{{\sqrt 5 }}{2}$

$\begin{array}{l}
d)\cos \alpha  = \dfrac{{ - 1}}{4}\\
Do:\dfrac{\pi }{2} < \alpha  < \pi  \Rightarrow \sin \alpha  > 0\\
 \Rightarrow \sin \alpha  = \sqrt {1 - {{\cos }^2}\alpha }  = \dfrac{{\sqrt {15} }}{4}\\
 \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} =  - \sqrt {15} ;\cot \alpha  = \dfrac{1}{{\tan \alpha }} = \dfrac{{ - 1}}{{\sqrt {15} }}
\end{array}$

Vậy $\sin \alpha  = \dfrac{{\sqrt {15} }}{4};\tan \alpha  =  - \sqrt {15} ;\cot \alpha  = \dfrac{{ - 1}}{{\sqrt {15} }}$

B2:

$\begin{array}{l}
a)A = \dfrac{{2\sin 2\alpha  - \sin 4\alpha }}{{2\sin 2\alpha  + \sin 4\alpha }}\\
 = \dfrac{{2\sin 2\alpha  - 2\sin 2\alpha \cos 2\alpha }}{{2\sin 2\alpha  + 2\sin 2\alpha \cos 2\alpha }}\\
 = \dfrac{{2\sin 2\alpha \left( {1 - \cos 2\alpha } \right)}}{{2\sin 2\alpha \left( {1 + \cos 2\alpha } \right)}}\\
 = \dfrac{{1 - \cos 2\alpha }}{{1 + \cos 2\alpha }}\\
 = \dfrac{{2{{\sin }^2}\alpha }}{{2{{\cos }^2}\alpha }}\\
 = {\tan ^2}\alpha 
\end{array}$

Bạn kiểm tra lại đề câu $b$

$\begin{array}{l}
b)B = \dfrac{{\cos a + 2\cos 2a + \cos 3a}}{{\sin a + 2\sin 2a + \sin 3a}}\\
 = \dfrac{{\left( {\cos a + \cos 3a} \right) + 2\cos 2a}}{{\left( {\sin a + \sin 3a} \right) + 2\sin 2a}}\\
 = \dfrac{{2\cos 2a\cos a + 2\cos 2a}}{{2\sin 2a\cos a + 2\sin 2a}}\\
 = \dfrac{{2\cos 2a\left( {\cos a + 1} \right)}}{{2\sin 2a\left( {\cos a + 1} \right)}}\\
 = \cot 2a
\end{array}$