Giải thích các bước giải:
a.Ta có :
$\cot \alpha +\tan\alpha=\dfrac{\cos\alpha}{\sin\alpha}+\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos\alpha.\sin\alpha}=\dfrac{1}{\cos\alpha.\sin\alpha}$
$\cot \alpha -\tan\alpha=\dfrac{\cos\alpha}{\sin\alpha}-\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos\alpha.\sin\alpha}=\dfrac{1-2\sin^2\alpha}{\cos\alpha.\sin\alpha}$
$\rightarrow A=\dfrac{\dfrac{1}{\cos\alpha.\sin\alpha}}{\dfrac{1-2\sin^2\alpha}{\cos\alpha.\sin\alpha}}=\dfrac{1}{1-2\sin^2\alpha}=\dfrac{25}{7}$
c.Ta có : $\cot\alpha=-3\rightarrow \tan\alpha=\dfrac{-1}{3}$
$\rightarrow C=\dfrac{(\sin^2\alpha+2\sin\alpha.\cos\alpha-2\cos^2\alpha):\sin^2\alpha}{(2\sin^2\alpha-3\sin\alpha.\cos\alpha+4\cos^2\alpha):\sin^2\alpha }$
$\rightarrow C=\dfrac{1+2.\dfrac{\cos\alpha}{\sin\alpha}-2.\dfrac{\cos^2\alpha}{\sin^2\alpha}}{2-3.\dfrac{\cos\alpha}{\sin\alpha}+4\dfrac{\cos^2\alpha}{\sin^2\alpha} }$
$\rightarrow C=\dfrac{1+2.\cot \alpha-2\cot^2 \alpha}{2-3.\cot \alpha+4\cot^2 \alpha}$
$\rightarrow C=-1$
