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Ta có:
$\frac{1}{\sin A}+\cot A=\frac{1+\cos A}{\sin A}=\frac{1}{\displaystyle\tan \frac{A}{2}}$
$\Rightarrow \tan\frac{A}{2}=\frac{c-b}{a}\Rightarrow c>b$
Từ đó ta có:
$\frac{b^2+c^2-a^2}{2bc}=\cos A=\frac{\displaystyle1-\tan^2\frac{A}{2}}{\displaystyle1+\tan^2\frac{A}{2}}=\frac{\displaystyle1-\frac{(c-b)^2}{a^2}}{\displaystyle1+\frac{(c-b)^2}{a^2}}$
$\Rightarrow \frac{(b^2-c^2)^2-a^4}{bc(a^2+(b-c)^2)}=0$
$\Leftrightarrow a^4=(b^2-c^2)^2$
$\Leftrightarrow a^2=c^2-b^2\Leftrightarrow \Delta ABC$ vuông tại $C$.
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