\(a.\left(c.cosC-b.cosB\right)=a.\left(c.\dfrac{a^2+b^2-c^2}{2ab}-b.\dfrac{a^2+c^2-b^2}{3ac}\right)\)

\(=\dfrac{\left(a^2+b^2-c^2\right)c^2}{2bc}-\dfrac{\left(a^2+c^2-b^2\right)b^2}{2bc}\)

\(=\dfrac{\left(b^2-c^2\right)\left(b^2+c^2-a^2\right)}{2bc}=\left(b^2-c^2\right)cosA\)

\(a\left(c.cosC-b.cosB\right)=a\left(c.\dfrac{a^2+b^2-c^2}{2ab}-b.\dfrac{a^2+c^2-b^2}{2ac}\right)\)

\(=\dfrac{\left(a^2+b^2-c^2\right).c^2}{2bc}-\dfrac{\left(a^2+c^2-b^2\right).b^2}{2bc}\)

\(=\dfrac{b^4-c^4+a^2c^2-a^2b^2}{2bc}\)

\(=\dfrac{\left(b^2-c^2\right)\left(b^2+c^2-a^2\right)}{2bc}=\left(b^2-c^2\right).cosA\)

Theo hệ quả định lí cô sin trong tam giác ta có:  cosB =   c 2 + ​ a 2 − b 2 2 c a

Từ  giả thiết: c = a. cosB nên: 

c = a . c 2 + a 2 − b 2 2. c a ⇒ c = c 2 + a 2 − b 2 2 c ⇒ 2 c 2 = c 2 + a 2 − b 2 ⇒ a 2 = b 2 + c 2

Do đó, tam giác ABC vuông tại A.

ĐÁP ÁN C

a/ \(b^2-c^2=ab.cosC-ac.cosB\)

Ta có: \(b.cosC-c.cosB=ab.\dfrac{a^2+b^2-c^2}{2ab}-ac.\dfrac{a^2+c^2-b^2}{2ac}\)

\(=\dfrac{a^2+b^2-c^2}{2}-\dfrac{a^2+c^2-b^2}{2}=\dfrac{2b^2-2c^2}{2}=b^2-c^2\) (đpcm)

b/ \(ac.cosC-ab.cosB=ac.\dfrac{a^2+b^2-c^2}{2ab}-ab.\dfrac{a^2+c^2-b^2}{2ac}\)

\(=\dfrac{c^2\left(a^2+b^2-c^2\right)-b^2\left(a^2+c^2-b^2\right)}{2bc}=\dfrac{\left(ac\right)^2-\left(ab\right)^2+b^4-c^4}{2bc}\)

\(=\dfrac{-a^2\left(b^2-c^2\right)+\left(b^2-c^2\right)\left(b^2+c^2\right)}{2bc}=\left(b^2-c^2\right).\dfrac{\left(b^2+c^2-a^2\right)}{2bc}\)

\(=\left(b^2-c^2\right).cosA\) (đpcm)

c/ \(cotA+cotB+cotC=\dfrac{cosA}{sinA}+\dfrac{cosB}{sinB}+\dfrac{cosC}{sinC}=\dfrac{2R.cosA}{a}+\dfrac{2R.cosB}{b}+\dfrac{2R.cosC}{c}\)

\(=2R\left(\dfrac{b^2+c^2-a^2}{2abc}+\dfrac{a^2+c^2-b^2}{2abc}+\dfrac{a^2+b^2-c^2}{2abc}\right)\)

\(=2R\left(\dfrac{a^2+b^2+c^2}{2abc}\right)=\dfrac{a^2+b^2+c^2}{abc}.R\) (đpcm)

a: \(a^4=b^4+c^4\)

=>\(a^4>c^4\) và \(a^4>b^4\)

=>a>c và a>b

=>a là cạnh lớn nhất

=>\(\hat{BAC}\) là góc lớn nhất

\(a^4=b^4+c^4\)

=>\(a^4 = (b^2+c^2)^2 - 2b^2c^2 < (b^2+c^2)^2 \Rightarrow a^2 < b^2 + c^2\)

=>\(b^2+c^2-a^2>0\)

=>\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\) >0

=>\(\hat{A}<90^0\)

mà \(\hat{BAC}\) đã là góc lớn nhất trong ΔABC

nên \(\hat{ABC}<90^0;\hat{ACB}<90^0\)

b: Xét ΔABC có \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\)

=>\(\sin A=\frac{a}{2R}\) ; \(\sin B = \frac{b}{2R}, \quad \sin C = \frac{c}{2R}\)

Xét ΔABC có \(\cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab}\)

\(\tan B \cdot \tan C = \frac{\sin B \cdot \sin C}{\cos B \cdot \cos C}\)

\(=\frac{\frac{b}{2R} \cdot\frac{c}{2R}}{\frac{a^2 + c^2 - b^2}{2ac} \cdot\frac{a^2 + b^2 - c^2}{2ab}}\)

\(=\frac{\frac{bc}{4R^2}}{\frac{(a^2 + c^2 - b^2)(a^2 + b^2 - c^2)}{4a^2bc}}=\frac{a^2 b^2 c^2}{R^2 (a^2 + c^2 - b^2)(a^2 + b^2 - c^2)}\)

\(=\frac{a^2b^2c^2}{R^2\cdot\left\lbrack(a^2+c^2-b^2)(a^2+b^2-c^2)\right\rbrack}\)

\(=\frac{a^2b^2c^2}{R^2\left\lbrace\left[a^2+(c^2-b^2)\right]\left[a^2-(c^2-b^2)\right]\right\rbrace}\)

\(=\frac{a^2b^2c^2}{R^2\cdot\left\lbrack a^4-\left(c^2-b^2\right)^2\right\rbrack}=\frac{a^2b^2c^2}{R^2\cdot\left\lbrack a^4-(c^4-2b^2c^2+b^4)\right\rbrack}\)

\(=\frac{a^2b^2c^2}{R^2\cdot\left\lbrack a^4-(b^4+c^4)+2b^2c^2\right\rbrack}=\frac{a^2b^2c^2}{R^2\cdot2b^2c^2}=\frac{a^2}{2R^2}=2\cdot\sin^2A\)

\(\dfrac{b^2-a^2}{2c}=b.\dfrac{\left(b^2+c^2-a^2\right)}{2bc}-a.\dfrac{\left(a^2+c^2-b^2\right)}{2ac}\)

\(\Leftrightarrow\dfrac{b^2-a^2}{2c}=\dfrac{b^2+c^2-a^2}{2c}-\dfrac{a^2+c^2-b^2}{2c}\)

\(\Leftrightarrow b^2-a^2=\left(b^2+c^2-a^2\right)-\left(a^2+c^2-b^2\right)\)

\(\Leftrightarrow3b^2=3a^2\Leftrightarrow a=b\)

Hay tam giác cân tại C

Áp dụng hệ quả của định lí sin và định lí cosin, ta có:

\(\frac{a}{{\sin A}} = 2R \Rightarrow \sin A = \frac{a}{{2R}}\)

và \(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)

\( \Rightarrow \cot A = \frac{{\cos A}}{{\sin A}} = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}:\frac{a}{{2R}} = R.\frac{{{b^2} + {c^2} - {a^2}}}{{abc}}\)

Tương tự ta có: \(\cot B = R.\frac{{{a^2} + {c^2} - {b^2}}}{{abc}}\) và \(\cot C = R.\frac{{{a^2} + {b^2} - {c^2}}}{{abc}}\)

\(\begin{array}{l} \Rightarrow \cot A + \cot B + \cot C = \frac{R}{{abc}}\left[ {\left( {{b^2} + {c^2} - {a^2}} \right) + \left( {{a^2} + {c^2} - {b^2}} \right) + \left( {{a^2} + {b^2} - {c^2}} \right)} \right]\\ = \frac{R}{{abc}}\left( {2{b^2} + 2{c^2} + 2{a^2} - {a^2} - {c^2} - {b^2}} \right) = \frac{{R({a^2} + {b^2} + {c^2})}}{{abc}}\end{array}\)