Giải:

a) Ta có $\cos^2\alpha = 1 - \sin^2\alpha = 1 - \left(\dfrac{1}{3}\right)^2 = \dfrac{8}{9}$.
Vì $-\dfrac{\pi}{2} < \alpha < 0$ nên $\cos\alpha > 0$. Do đó $\cos\alpha = \dfrac{2\sqrt{2}}{3}.$

Suy ra $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{1/3}{2\sqrt{2}/3} = \dfrac{1}{2\sqrt{2}}$, và $\cot\alpha = 2\sqrt{2}.$

b) $\sin^2\alpha = 1 - \cos^2\alpha = 1 - (-0{,}7)^2 = \dfrac{51}{100}.$
Vì $\pi < \alpha < \dfrac{3\pi}{2}$ nên $\sin\alpha < 0$. Do đó $\sin\alpha = -\dfrac{\sqrt{51}}{10}.$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{-\sqrt{51}/10}{-0{,}7} = \dfrac{\sqrt{51}}{7}$, và $\cot\alpha = \dfrac{7}{\sqrt{51}}.$

c) $\cot^2\alpha = \dfrac{1}{\tan^2\alpha} = \dfrac{1}{4}.$
$\sin^2\alpha = \dfrac{\tan^2\alpha}{1 + \tan^2\alpha} = \dfrac{4}{5}, \quad \cos^2\alpha = \dfrac{1}{5}.$
Vì $0 < \alpha < \dfrac{\pi}{2}$ nên $\sin\alpha, \cos\alpha > 0$. Do đó $\sin\alpha = \dfrac{2\sqrt{5}}{5}, \ \cos\alpha = \dfrac{\sqrt{5}}{5}.$

d) $\tan\alpha = \dfrac{1}{\cot\alpha} = \dfrac{3}{7}$.
$\sin^2\alpha = \dfrac{\tan^2\alpha}{1 + \tan^2\alpha} = \dfrac{9}{58}$, $\cos^2\alpha = \dfrac{1}{1 + \tan^2\alpha} = \dfrac{49}{58}.$
Vì $\pi < \alpha < \dfrac{3\pi}{2}$ nên $\sin\alpha, \cos\alpha < 0$. Do đó $\sin\alpha = -\dfrac{3\sqrt{58}}{58}, \ \cos\alpha = -\dfrac{7\sqrt{58}}{58}.$

Giải:

a) $\dfrac{\sin^3\alpha + \cos^3\alpha}{\sin\alpha + \cos\alpha} = \dfrac{(\sin\alpha + \cos\alpha)(\sin^2\alpha - \sin\alpha\cos\alpha + \cos^2\alpha)}{\sin\alpha + \cos\alpha} = 1 - \sin\alpha\cos\alpha.$

b) $\dfrac{1+\tan\alpha}{1-\tan\alpha} + \dfrac{1+\cot\alpha}{1-\cot\alpha} = \dfrac{1+\tan\alpha}{1-\tan\alpha} + \dfrac{\cot\alpha(\cot\alpha+1)}{\cot\alpha-1} = \dfrac{1+\tan\alpha}{1-\tan\alpha} + \dfrac{1+\frac{1}{\tan\alpha}}{1-\frac{1}{\tan\alpha}} = 0.$

Giải:

a) Ta có: $\sin(3\pi - \alpha) = \sin(\pi + 2\pi - \alpha) = \sin(\pi - \alpha) = \sin\alpha.$
$\sin\left(\alpha + \dfrac{5\pi}{2}\right) = \sin\left(\alpha + 2\pi + \dfrac{\pi}{2}\right) = \cos\alpha.$
Suy ra $\sin^2(3\pi - \alpha) + \sin^2\left(\alpha + \dfrac{5\pi}{2}\right) = \sin^2\alpha + \cos^2\alpha = 1.$

b) $\tan(-\alpha + 11\pi) = \tan(-\alpha + \pi) = \tan(-\alpha) = -\tan\alpha.$
$\sin\left(\alpha - \dfrac{3\pi}{2}\right) = \cos\alpha.$
Suy ra: $$[1 + \tan^2(-\alpha + 11\pi)] \cdot \sin^2\left(\alpha - \dfrac{3\pi}{2}\right) = (1 + \tan^2\alpha)\cos^2\alpha = 1.$$