Giúp em với ạ em cảm ơn

Vì $\pi < \alpha < \dfrac{3\pi}{2}$ $($ góc phần tư thứ ba$)$

$\Rightarrow \sin \alpha < 0$ và $\cos \alpha < 0$

Ta có:

$\dfrac{1}{\cos^2 \alpha} = 1 + \tan^2 \alpha = 1 + (\sqrt{5})^2 = 6$

$\Rightarrow \cos^2 \alpha = \dfrac{1}{6}$

$\Rightarrow \cos \alpha = -\dfrac{1}{\sqrt{6}} = -\dfrac{\sqrt{6}}{6}$ (do $\cos \alpha < 0$)

$\Rightarrow \sin \alpha = \tan \alpha \cdot \cos \alpha = \sqrt{5} \cdot \left(-\dfrac{\sqrt{6}}{6}\right) = -\dfrac{\sqrt{30}}{6}$

Xét: $\sin\left(\alpha + \dfrac{\pi}{3}\right)$

$= \sin \alpha \cos\dfrac{\pi}{3} + \cos \alpha \sin\dfrac{\pi}{3}$

$= \left(-\dfrac{\sqrt{30}}{6}\right) \cdot \dfrac{1}{2} + \left(-\dfrac{\sqrt{6}}{6}\right) \cdot \dfrac{\sqrt{3}}{2}$

$= -\dfrac{\sqrt{30}}{12} - \dfrac{\sqrt{18}}{12}$

$= -\dfrac{\sqrt{30} + 3\sqrt{2}}{12}$

Xét: $\cos\left(\alpha - \dfrac{\pi}{6}\right)$

$= \cos \alpha \cos\dfrac{\pi}{6} + \sin \alpha \sin\dfrac{\pi}{6}$

$= \left(-\dfrac{\sqrt{6}}{6}\right) \cdot \dfrac{\sqrt{3}}{2} + \left(-\dfrac{\sqrt{30}}{6}\right) \cdot \dfrac{1}{2}$

$= -\dfrac{\sqrt{18}}{12} - \dfrac{\sqrt{30}}{12}$

$= -\dfrac{3\sqrt{2} + \sqrt{30}}{12}$

Xét: $\tan\left(\dfrac{\pi}{4} - \alpha\right)$

$= \dfrac{\tan\dfrac{\pi}{4} - \tan \alpha}{1 + \tan\dfrac{\pi}{4}\tan \alpha}$

$= \dfrac{1 - \sqrt{5}}{1 + 1 \cdot \sqrt{5}}$

$= \dfrac{1 - \sqrt{5}}{\sqrt{5} + 1}$

$= \dfrac{(1 - \sqrt{5})^2}{1 - 5}$

$= \dfrac{1 - 2\sqrt{5} + 5}{-4}$

$= \dfrac{6 - 2\sqrt{5}}{-4}$

$= \dfrac{\sqrt{5} - 3}{2}$

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