Câu 1. Ta có: $ sinAcosA\frac{cosC}{sinC}=\frac{sinA.sinC+cosA.cosC}{sinC}=\frac{cos(A-C)}{sinC}$
$\Rightarrow \frac{sinC}{cosB}=\frac{cos(A-C)}{sinC}$
$\Leftrightarrow sin^2C=cosB.cos(A-C)$ Mà $B=\pi-(A+C)$
$\Leftrightarrow sin^2C=-cos(A+C).cos(A-C)$
$\Leftrightarrow sin^2C=-(cosA.cosC+sinA.sinC)(cosA.cosC-sinA.sinC)$
$\Leftrightarrow sin^2C-sin^2A.sin^2C+cos^2A.cos^2C=0$
$\Leftrightarrow sin^2C(1-sin^2A)+cos^2A.cos^2C=0$
$\Leftrightarrow sin^2C.cos^2A+cos^2A.cos^2C=0$
$\Leftrightarrow cos^2A(sin^2C+cos^2C)=0$
$\Leftrightarrow cosA=0$
$\Rightarrow A=\frac{\pi}{2}$
Vậy $\Delta ABC$ vuông tại A