Chứng minh rằng tam giác ABC vuông khi b / cos B + c / cos C

Chứng minh rằng tam giác ABC vuông khi $\frac{b}{{\cos B}} + \frac{c}{{\cos C}} = \frac{a}{{\sin B.\sin C}}$.

$\frac{b}{{\cos B}} + \frac{c}{{\cos C}} = \frac{a}{{\sin B.\sin C}}$

⇔ $\frac{b}{{\frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}}} + \frac{c}{{\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}}} = \frac{{abc}}{{\sin B.\sin C.bc}}$

⇔ $\frac{{2abc}}{{{a^2} + {c^2} - {b^2}}} + \frac{{2abc}}{{{a^2} + {b^2} - {c^2}}} = \frac{a}{{bc}}.\frac{b}{{\sin B}}.\frac{c}{{\sin C}}$

⇔ $\frac{{2abc}}{{{a^2} + {c^2} - {b^2}}} + \frac{{2abc}}{{{a^2} + {b^2} - {c^2}}} = \frac{{4a{R^2}}}{{bc}}$

⇔ $\frac{{4{a^3}bc}}{{\left( {{a^2} + {c^2} - {b^2}} \right)\left( {{a^2} + {b^2} - {c^2}} \right)}} = \frac{{4a{R^2}}}{{bc}}$

⇔ $\frac{{{a^2}bc}}{{\left( {{a^2} + {c^2} - {b^2}} \right)\left( {{a^2} + {b^2} - {c^2}} \right)}} = \frac{{{R^2}}}{{bc}}$

⇔ R²(a² + c² – b²)(a² + b² – c²) = (abc)²

⇔ $\frac{{\left( {{a^2} + {c^2} - {b^2}} \right).R}}{{abc}}.\frac{{\left( {{a^2} + {b^2} - {c^2}} \right).R}}{{abc}} = 1$

⇔$\frac{{\left( {{a^2} + {c^2} - {b^2}} \right)}}{{2ac}}.\frac{{2R}}{b}.\frac{{\left( {{a^2} + {b^2} - {c^2}} \right)}}{{2ab}}.\frac{{2R}}{c} = 1$

Mà $\frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R$

Suy ra: $\frac{{\left( {{a^2} + {c^2} - {b^2}} \right)}}{{2ac}}.\frac{{2R}}{b}.\frac{{\left( {{a^2} + {b^2} - {c^2}} \right)}}{{2ab}}.\frac{{2R}}{c} = 1$

⇔ $\frac{{\cos B}}{{\sin B}}.\frac{{\cos C}}{{\sin C}} = 1$

⇔ cotB.cotC = 1

⇔ cotB = $\frac{1}{{\cot C}} = \tan C$

Suy ra: tam giác ABC vuông vì khi góc $\widehat B,\widehat C$phụ nhau thì tan góc này bằng cotan góc kia.

Vậy tam giác ABC vuông khi $\frac{b}{{\cos B}} + \frac{c}{{\cos C}} = \frac{a}{{\sin B.\sin C}}$.