Giải thích các bước giải:

b.Ta có:

$180^o<\alpha <270^o$

$\to \sin\alpha<0,\cos\alpha<0,\tan\alpha>0, \cot\alpha>0$

$\to \cos\alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-(-\dfrac13)^2}=-\dfrac{2\sqrt{2}}{3}$

     $\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=-\dfrac13:(-\dfrac{2\sqrt{2}}{3})=\dfrac{1}{2\sqrt{2}}$

     $\cot\alpha=\dfrac1{\tan\alpha}=2\sqrt2$

d.Ta có: $\pi<\alpha<\dfrac32\pi\to \sin\alpha<0,\cos\alpha<0,\tan\alpha>0,\cot\alpha>0$

            $\cot\alpha=3$

$\to \tan\alpha=\dfrac1{\cot\alpha}=\dfrac13$

Ta có:

$\cot\alpha=3$

$\to \dfrac{\cos\alpha}{\sin\alpha}=3$

$\to \dfrac{\cos^2\alpha}{\sin^2\alpha}=9$

$\to \dfrac{\cos^2\alpha}{\sin^2\alpha}+1=10$

$\to \dfrac{\cos^2\alpha+\sin^2\alpha}{\sin^2\alpha}=10$

$\to \dfrac1{\sin^2\alpha}=10$

$\to \sin^2\alpha=\dfrac1{10}\to \cos^2\alpha=1-\sin^2\alpha=\dfrac9{10}$

$\to \sin\alpha=-\dfrac1{\sqrt{10}}, \cos\alpha=\dfrac{-3}{\sqrt{10}}$

f.Ta có:

$\dfrac{2025\pi}2<\alpha<\dfrac{2027\pi}2$

 $\to 1012\pi+\dfrac12\pi<\alpha<1012\pi+\dfrac32\pi$

$\to \cos\alpha<0$

Ta có:

$\tan\alpha=-\dfrac43\to \cot\alpha=\dfrac1{\tan\alpha}=-\dfrac34$

$\tan^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}$

$\to \dfrac{16}9=\dfrac{\sin^2\alpha}{\cos^2\alpha}$

$\to \dfrac{16}9+1=\dfrac{\sin^2\alpha}{\cos^2\alpha}+1$

$\to \dfrac{25}9=\dfrac1{\cos^2\alpha}$

$to \cos^2\alpha=\dfrac9{25}$

$\to \cos\alpha=\dfrac{-3}{5}$ vì $\cos\alpha<0$

Lại có:

$\tan\alpha=-\dfrac43$

$\to \dfrac{\sin\alpha}{\cos\alpha}=-\dfrac43$

$\to \sin\alpha=-\dfrac43\cos\alpha=\dfrac{-4}3\cdot \dfrac{-3}5=\dfrac45$