Question

Trong tam giác ABC , nếu \(\frac{tanA}{tanC}=\frac{sin^2A}{sin^2C}\)thì tam giác ABC là tam giác gì?

Answer 1

\(tanA.cotC-\frac{sin^2A}{sin^2C}=0\)

\(\Leftrightarrow\frac{sinA.cosC}{cosA.sinC}-\frac{sin^2A}{sin^2C}=0\)

\(\Leftrightarrow\frac{sinA}{sinC}\left(\frac{cosC}{cosA}-\frac{sinA}{sinC}\right)=0\)

\(\Leftrightarrow\frac{cosC}{cosA}-\frac{sinA}{sinC}=0\) (do \(0< A< 180\Rightarrow\frac{sinA}{sinC}\ne0\))

\(\Rightarrow sinC.cosC=sinA.cosA\)

\(\Leftrightarrow\frac{1}{2}sin2C=\frac{1}{2}sin2A\)

\(\Rightarrow sin2C=sin2A\Rightarrow\left[{}\begin{matrix}2C=2A\\2C=180^0-2A\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}C=A\\A+C=90^0\end{matrix}\right.\) \(\Rightarrow\Delta ABC\) vuông hoặc cân tại B