b + c = 2a

⇔ \(\dfrac{b+c}{2R}=\dfrac{2a}{2R}\) (1) với R là bán kính đường tròn ngoại tiếp

Theo định lí sin \(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=2R\)

nên (1) ⇔ sinB + sinC = 2sinA

Chọn B

TL:

sinA+sinB+sinC=1-cosA+cosB+cosC => Tam giác ABC Vuông tại A

Vế trái = sinA + sinB + sinC

= 2sin(A + B)/2.cos(A - B)/2 + 2sinC/2.cosC/2

= 2cosC/2.cos(A - B)/2 + 2sinC/2.cosC/2

= 2cosC/2[cos(A - B)/2 + sinC/2]

=2.cosC/2.[cos(A - B)/2 + cos(A + B)/2]

= 4.cosC/2.cosB/2.cosA/2

Vế phải = 1 - cosA + cosB + cosC

= 2sin²A/2 + 2cos(B + C)/2.cos(B - C)/2

= 2.sinA/2[sinA/2 + cos(B - C)/2] (vì cos(B + C)/2 = sinA/2)

= 2.sinA/2[cos(B + C)/2 + cos(B - C)/2

= 4.sinA/2.cosB/2.cosC/2

Vậy sinA + sinB + sinC = 1 - cosA + cosB + cosC

<=> cosA/2.cosB/2.cosC/2 = sinA/2.cosB/2.cosC/2

<=> cosB/2.cosC/2(sinA/2 - cosA/2) = 0

mà cosB/2 ≠ 0 và cosC/2 ≠ 0

=> sinA/2 = cosA/2

<=> A/2 = 45^o

<=> A = 90^o

tam giác ABC vuông tại A

\(a,cosC=\dfrac{5}{13}\\ Ta,có:cos^2C+sin^2C=1\\ \Rightarrow sinC=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\\ cosB+sinC=1\\ \Leftrightarrow cosB+\dfrac{12}{13}=1\\ \Rightarrow cosB=\dfrac{1}{13}\\ tanC=\dfrac{sinC}{cosC}=\dfrac{\dfrac{12}{13}}{\dfrac{5}{13}}=\dfrac{12}{5}\)

\(b,tanB=\dfrac{1}{5}\Rightarrow\dfrac{sinB}{cosB}=\dfrac{1}{5}\Rightarrow cosB=5sinB\\ E=\dfrac{sinB-3cosB}{2sinB+3cosB}=\dfrac{sinB-3.5.sinB}{2sinB+3.5.sinB}=\dfrac{-14sinB}{17sinB}=-\dfrac{14}{17}\)

C

Lời giải:

Áp dụng 1 số công thức lượng giác:
\(\sin A=\frac{\sin B+\sin C}{\cos B+\cos C}=\frac{2\sin (\frac{B+C}{2})\cos (\frac{B-C}{2})}{2\cos (\frac{B+C}{2})\cos (\frac{B-C}{2})}=\frac{\sin \frac{B+C}{2}}{\cos \frac{B+C}{2}}\)

\(=\tan \frac{B+C}{2}=\tan (\frac{\pi-A}{2})=\cot \frac{A}{2}\)

\(\Leftrightarrow 2\sin \frac{A}{2}\cos \frac{A}{2}=\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\) (trong tam giác, \(\widehat{A}\neq 0\rightarrow \sin \frac{A}{2}\neq 0)\)

\(\Leftrightarrow \cos \frac{A}{2}(2\sin^2 \frac{A}{2}-1)=0\)

\(\Rightarrow \left[\begin{matrix} \cos \frac{A}{2}=0\rightarrow \frac{\widehat{A}}{2}=\frac{\pi}{2}\rightarrow \widehat{A}=\pi (\text{vô lý})\\ \sin \frac{A}{2}=\frac{1}{\sqrt{2}}\rightarrow \frac{\widehat{A}}{2}=\frac{\pi}{4}\rightarrow \widehat{A}=\frac{1}{2}\pi=90^0 \end{matrix}\right.\)

Do đó tam giác ABC vuông tại A

a)\(VT=sinA+sinB+sinC=2sin\frac{A+B}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)(đpcm)

b)Ta có:\(A+B+C=180^O\)

\(\Rightarrow tan\left(A+B\right)=tan\left(-C\right)=-tanC\)

\(\Leftrightarrow\frac{tanA+tanB}{1-tanA.tanB}=-tanC\Leftrightarrow tanA+tanB+tanC=tanA.tanB.tanC\left(đpcm\right)\)