Giúp em với mọi người ơi em đang cần gấp huhu

a
Vì $90^\circ < \alpha < 180^\circ$ nên $\cos \alpha < 0$.
Ta có $\sin^2 \alpha + \cos^2 \alpha = 1$
$\Leftrightarrow \left(\dfrac{1}{3}\right)^2 + \cos^2 \alpha = 1$
$\Leftrightarrow \dfrac{1}{9} + \cos^2 \alpha = 1$
$\Leftrightarrow \cos^2 \alpha = 1 - \dfrac{1}{9}$
$\Leftrightarrow \cos^2 \alpha = \dfrac{8}{9}$
$\Leftrightarrow \cos \alpha = -\sqrt{\dfrac{8}{9}}$ (vì $\cos \alpha < 0$)
$\Leftrightarrow \cos \alpha = -\dfrac{2\sqrt{2}}{3}$
$\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha} = \dfrac{\dfrac{1}{3}}{-\dfrac{2\sqrt{2}}{3}} = -\dfrac{1}{2\sqrt{2}} = -\dfrac{\sqrt{2}}{4}$
b
Vì $0^\circ < x < 180^\circ$ nên $\sin x > 0$.
Ta có $\sin^2 x + \cos^2 x = 1$
$\Leftrightarrow \sin^2 x + \left(-\dfrac{2}{3}\right)^2 = 1$
$\Leftrightarrow \sin^2 x + \dfrac{4}{9} = 1$
$\Leftrightarrow \sin^2 x = 1 - \dfrac{4}{9}$
$\Leftrightarrow \sin^2 x = \dfrac{5}{9}$
$\Leftrightarrow \sin x = \sqrt{\dfrac{5}{9}}$ (vì $\sin x > 0$)
$\Leftrightarrow \sin x = \dfrac{\sqrt{5}}{3}$
$\cot x = \dfrac{\cos x}{\sin x} = \dfrac{-\dfrac{2}{3}}{\dfrac{\sqrt{5}}{3}} = -\dfrac{2}{\sqrt{5}} = -\dfrac{2\sqrt{5}}{5}$
c
Vì $\tan \alpha = -\sqrt{2} < 0$ và $0^\circ < \alpha < 180^\circ$ nên $90^\circ < \alpha < 180^\circ$.
Khi đó $\sin \alpha > 0$ và $\cos \alpha < 0$.
Ta có $1 + \tan^2 \alpha = \dfrac{1}{\cos^2 \alpha}$
$\Leftrightarrow 1 + (-\sqrt{2})^2 = \dfrac{1}{\cos^2 \alpha}$
$\Leftrightarrow 1 + 2 = \dfrac{1}{\cos^2 \alpha}$
$\Leftrightarrow 3 = \dfrac{1}{\cos^2 \alpha}$
$\Leftrightarrow \cos^2 \alpha = \dfrac{1}{3}$
$\Leftrightarrow \cos \alpha = -\sqrt{\dfrac{1}{3}}$ (vì $\cos \alpha < 0$)
$\Leftrightarrow \cos \alpha = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}$
Ta có $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha}$
$\Leftrightarrow \sin \alpha = \tan \alpha \cdot \cos \alpha$
$\Leftrightarrow \sin \alpha = (-\sqrt{2}) \cdot \left(-\dfrac{\sqrt{3}}{3}\right)$
$\Leftrightarrow \sin \alpha = \dfrac{\sqrt{6}}{3}$
$\cot \alpha = \dfrac{1}{\tan \alpha} = \dfrac{1}{-\sqrt{2}} = -\dfrac{\sqrt{2}}{2}$
d
Vì $\cot \alpha = 3 > 0$ và $0^\circ < \alpha < 180^\circ$ nên $0^\circ < \alpha < 90^\circ$.
Khi đó $\sin \alpha > 0$ và $\cos \alpha > 0$.
Ta có $1 + \cot^2 \alpha = \dfrac{1}{\sin^2 \alpha}$
$\Leftrightarrow 1 + 3^2 = \dfrac{1}{\sin^2 \alpha}$
$\Leftrightarrow 1 + 9 = \dfrac{1}{\sin^2 \alpha}$
$\Leftrightarrow 10 = \dfrac{1}{\sin^2 \alpha}$
$\Leftrightarrow \sin^2 \alpha = \dfrac{1}{10}$
$\Leftrightarrow \sin \alpha = \sqrt{\dfrac{1}{10}}$ (vì $\sin \alpha > 0$)
$\Leftrightarrow \sin \alpha = \dfrac{1}{\sqrt{10}} = \dfrac{\sqrt{10}}{10}$
Ta có $\cot \alpha = \dfrac{\cos \alpha}{\sin \alpha}$
$\Leftrightarrow \cos \alpha = \cot \alpha \cdot \sin \alpha$
$\Leftrightarrow \cos \alpha = 3 \cdot \dfrac{\sqrt{10}}{10}$
$\Leftrightarrow \cos \alpha = \dfrac{3\sqrt{10}}{10}$
$\tan \alpha = \dfrac{1}{\cot \alpha} = \dfrac{1}{3}$