Đáp án:
Bài 1:
$\tan \alpha =3\Rightarrow \dfrac{\sin \alpha }{\cos \alpha }=3\Rightarrow \sin \alpha =3\cos \alpha $
$A=\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha }=\dfrac{3\cos \alpha +\cos \alpha }{3\cos \alpha -\cos \alpha }=\dfrac{4\cos \alpha }{2\cos \alpha }=2$
Bài 2:
a)$VT=1+{{\tan }^{2}}\alpha =1+\dfrac{{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }=\dfrac{{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }=\dfrac{1}{{{\cos }^{2}}\alpha }=VP$
b)$VT=1+{{\cot }^{2}}\alpha =1+\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }=\dfrac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }=\dfrac{1}{{{\sin }^{2}}\alpha }=VP$
Bài 3:
$\cos B=\dfrac{AB}{BC}\Rightarrow BC=\dfrac{AB}{\cos B}=\dfrac{8}{\cos 60{}^\circ }=\dfrac{8}{\dfrac{1}{2}}=16\,\left( cm \right)$
$B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\Rightarrow AC=\sqrt{B{{C}^{2}}-A{{B}^{2}}}=\sqrt{{{16}^{2}}-{{8}^{2}}}=8\sqrt{3}\,\left( cm \right)$
