a) Xét ΔABE  và ΔACF có

Alà góc chung

AEB=AFC(=90^O)

=> ΔABE đồng dạng ΔACF (g.g)

=>AF/AE​=AC/AB​

=> AB/AE​=AC/AF​

XétΔAEF và  ΔABC có

AB/AE​=AC/AF​

Và Agóc chung

Suy raΔAEF đồng dạngΔABC( c.g.c) 

d: \(\sin B\cdot cosC+\sin C\cdot cosB\)

\(=\sin\left(B+C\right)\)

\(=\sin\left(180^0-A\right)=\sin A\)

e: tan A+tan B+tan C

\(=\tan\left(A+B\right)\left(1-\tan A\cdot\tan B\right)+\tan C\)

\(=\tan\left(180^0-C\right)\left(1-\tan A\cdot\tan B\right)+\tan C\)

\(=-\tan C\left(1-\tan A\cdot\tan B\right)+\tan C\)

\(=-\tan C+\tan A\cdot\tan B\cdot\tan C+\tan C=\tan A\cdot\tan B\cdot\tan C\)

Bài 2:

\(P = \cos A + \cos B + \cos C = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) + \cos C\)

\(\hat{A}+\hat{B}+\hat{C}=180^0\)

=>\(\hat{A}+\hat{B}=180^0-\hat{C}\)

=>\(\frac12\left(\hat{A}+\hat{B}\right)=90^0-\frac12\cdot\hat{C}\)

=>\(\cos\left(\dfrac{A+B}{2}\right)=\sin\left(\dfrac{C}{2}\right);\cos C=1-2\sin^2\left(\dfrac{C}{2}\right)\)

=>\(P = 2 \sin\left(\frac{C}{2}\right) \cos\left(\frac{A-B}{2}\right) + 1 - 2\sin^2\left(\frac{C}{2}\right)\)

Ta có: \(\cos\left(\dfrac{A-B}{2}\right) \le 1\)

\(\sin\left(\dfrac{C}{2}\right) > 0\)

Do đó: \(P \le 2 \sin\left(\frac{C}{2}\right) \cdot 1 + 1 - 2\sin^2\left(\frac{C}{2}\right)\)

=>\(P \le -2\sin^2\left(\frac{C}{2}\right) + 2\sin\left(\frac{C}{2}\right) + 1\)

=>\(P \le \frac{3}{2} - 2\left(\sin^2\left(\frac{C}{2}\right) - \sin\left(\frac{C}{2}\right) + \frac{1}{4}\right)\)

=>\(P\le\frac{3}{2}-2\left(\sin\left(\frac{C}{2}\right)-\frac{1}{2}\right)^2\le\frac32\)

Dấu '=' xảy ra khi \(\hat{A}=\hat{B}=\hat{C}=\frac{180^0}{3}=60^0\)

Xét tam giác ABC nhọn có \(BC^2=AB^2+AC^2-2AB\cdot AC\cdot\cos\widehat{A}\)
\(\Rightarrow\cos\widehat{A}=\dfrac{AB^2+AC^2-BC^2}{2AB\cdot AC}=\dfrac{AB^2+AC^2-BC^2}{4\cdot\dfrac{1}{2}AB\cdot AC}=\dfrac{AB^2+AC^2-BC^2}{4S_{ABC}}\)

Cmtt: \(\left\{{}\begin{matrix}\cos\widehat{B}=\dfrac{AB^2+BC^2-AC^2}{4S_{ABC}}\\\cos\widehat{C}=\dfrac{AC^2+BC^2-AB^2}{4S_{ABC}}\end{matrix}\right.\)
\(\Rightarrow\cos\widehat{A}+\cos\widehat{B}+\cos\widehat{C}\\ =\dfrac{AB^2+AC^2-BC^2+AB^2+BC^2-AC^2+AC^2+BC^2-AB^2}{4S_{ABC}}\\ =\dfrac{AB^2+AC^2+BC62}{4S_{ABC}}\)