Chủ đề này có 1 trả lời

[hoanganhtuan96vn]

$1, \frac{cos(B-C)}{sinA + cos(B-C))}=tanB$
$2, \frac{sinA+cosB}{sinB+cosA}=tanA$
$3, sin\frac{B}{2}=\sqrt{\frac{a-c}{a.c}}$
$4, cos(B-C)=\frac{2bc}{a^{2}}$

[MIM]

$2, \frac{sinA+cosB}{sinB+cosA}=tanA$

$\frac{sinA+cosB}{sinB+cosA}=tanA$

$\Leftrightarrow sinA(sinB+cosA)=cosA(sinA+cosB)$

$\Leftrightarrow sinAsinB=cosAcosB$

$\Leftrightarrow cos(A+B)=0$

$\Leftrightarrow \widehat{A}+\widehat{B}=90^{\circ}$

$\Leftrightarrow \widehat{C}=90^{\circ}$

$4, cos(B-C)=\frac{2bc}{a^{2}}$

$cos(B-C)=\frac{2bc}{a^{2}}$

$\Leftrightarrow cos(B-C)=\frac{2sinBsinC}{sin^2A}\Leftrightarrow 2sin(B+C)cos(B+C)=\frac{4sinBsinC}{sinA}$

$\Leftrightarrow sin2B+sin2C=\frac{4sinBsinC}{sinA}\Leftrightarrow sinBcosB+sinCcosC=\frac{2sinBsinC}{sinA}$

$\Leftrightarrow sinBcosB+sinCcosC-\frac{sinBsinC}{sinA}-\frac{sinBsinC}{sinA}=0$

$\Leftrightarrow sinB(sinAcosB-sin(A+B))+sinC(sinAcosC-sin(A+C))=0$

$\Leftrightarrow -sinBsinBcosA-sinCsinCcosA=0$

$\Leftrightarrow cosA=0\Leftrightarrow \widehat{A}=90^{\circ}$