Giải thích các bước giải:
Ta có:
$\cot\alpha=\sqrt{3}+\sqrt{2}$
$\to \dfrac{\cos\alpha}{\sin\alpha}=\sqrt{3}+\sqrt{2}$
$\to \dfrac{\cos^2\alpha}{\sin^2\alpha}=\sqrt{3}+\sqrt{2}$
$\to \cos^2\alpha=(\sqrt{3}+\sqrt{2})\sin^2\alpha$
$\to (\sqrt{3}+\sqrt{2}+1)\cos^2\alpha=(\sqrt{3}+\sqrt{2})(\sin^2\alpha+\cos^2\alpha)$
$\to (\sqrt{3}+\sqrt{2}+1)\cos^2\alpha=(\sqrt{3}+\sqrt{2})$
$\to \cos^2\alpha=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}+1}$
$\to \cos\alpha=\sqrt{\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}+1}}$ vì $\cos\alpha>0$
Lại có:
$\dfrac{\cos\alpha}{\sin\alpha}=\sqrt{3}+\sqrt{2}$
$\to \sin\alpha=\dfrac{\cos\alpha}{\sqrt{3}+\sqrt{2}}$
$\to \sin\alpha=\dfrac{\sqrt{\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}+1}}}{\sqrt{3}+\sqrt{2}}$
Ta có:
$\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\sqrt{3}+\sqrt{2}}$
