Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 2{\cos ^2}x - 1 = 1 - 2{\sin ^2}x\\
\Rightarrow \left\{ \begin{array}{l}
{\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}\\
{\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}
\end{array} \right.\\
\cot \dfrac{B}{2} = \dfrac{{a + c}}{b}\\
\Leftrightarrow {\cot ^2}\dfrac{B}{2} = \dfrac{{{{\left( {a + c} \right)}^2}}}{{{b^2}}}\\
\Leftrightarrow \dfrac{{{{\cos }^2}\dfrac{B}{2}}}{{{{\sin }^2}\dfrac{B}{2}}} = \dfrac{{{{\left( {a + c} \right)}^2}}}{{{b^2}}}\\
\Leftrightarrow \dfrac{{\dfrac{{\cos B + 1}}{2}}}{{\dfrac{{1 - \cos B}}{2}}} = \dfrac{{{{\left( {a + c} \right)}^2}}}{{{b^2}}}\\
\Leftrightarrow \dfrac{{\cos B + 1}}{{1 - \cos B}} = \dfrac{{{{\left( {a + c} \right)}^2}}}{{{b^2}}}\\
\Leftrightarrow \dfrac{{\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} + 1}}{{1 - \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}}} = \dfrac{{{{\left( {a + c} \right)}^2}}}{{{b^2}}}\\
\Leftrightarrow \dfrac{{{a^2} + {c^2} - {b^2} + 2ac}}{{2ac - \left( {{a^2} + {c^2}} \right) + {b^2}}} = \dfrac{{{{\left( {a + c} \right)}^2}}}{{{b^2}}}\\
\Leftrightarrow \left[ {{{\left( {a + c} \right)}^2} - {b^2}} \right].{b^2} = {\left( {a + c} \right)^2}.\left[ {{b^2} - {{\left( {a - c} \right)}^2}} \right]\\
\Leftrightarrow {\left( {a + c} \right)^2}.{b^2} - {b^4} = {\left( {a + c} \right)^2}.{b^2} - {\left( {a + c} \right)^2}.{\left( {a - c} \right)^2}\\
\Leftrightarrow {b^4} = {\left( {a + c} \right)^2}.{\left( {a - c} \right)^2}\\
\Leftrightarrow {b^4} = {\left( {{a^2} - {c^2}} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
{b^2} = {a^2} - {c^2}\\
{b^2} = {c^2} - {a^2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{b^2} + {c^2} = {a^2}\\
{b^2} + {a^2} = {c^2}
\end{array} \right.
\end{array}\)
Vậy tam giác ABC là tam giác vuông.
