$A\,=\dfrac{\cos^2 17^\circ+2\cos^2 73^\circ}{\cot 65^\circ.\cot 25^\circ}-\sin ^2 17^\circ\\\quad =\dfrac{\cos^2 17^\circ+2\sin^2 17^\circ}{\cot 65^\circ.\tan 65^\circ}-\sin^2 17^\circ\\\quad =\dfrac{\cos ^2 17^\circ+\sin^1 17^\circ+\sin^2 17^\circ}{1}-\sin^2 17^\circ\\\quad =1+\sin^2 17\circ-\sin^2 17^\circ\\\quad =1$
Vậy $A=1$
$\sin\alpha=\dfrac{1}{4}\\\leftrightarrow \sin^2\alpha=\dfrac{1}{16}$
mà $\sin^2\alpha+\cos^2\alpha=1$
$\to \cos^2\alpha=\dfrac{15}{16}\\\leftrightarrow \cos\alpha=\dfrac{\sqrt{15}}{4}(\text{vì}\,\,\alpha\,\,\text{là góc nhọn})$
$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac{\sqrt{15}}{4}}{\dfrac{1}{4}}=\sqrt{15}$
$\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{\dfrac 1 4}{\dfrac{\sqrt{15}}{4}}=\dfrac{\sqrt{15}}{15}$
Vậy $\cos\alpha=\dfrac{\sqrt{15}}{4},\tan\alpha=\sqrt{15},\cot\alpha=\dfrac{\sqrt{15}}{15}$ với $\alpha$ là góc nhọn
