2.
\(A=\dfrac{\dfrac{sinx.cos^2x}{sin^3x}+\dfrac{cos^3x}{sin^3x}}{\dfrac{cos^3x}{sin^3x}-\dfrac{sin^3x}{sin^3x}}=\dfrac{cot^2x+cot^3x}{cot^3x-1}=\dfrac{\left(-3\right)^2+\left(-3\right)^3}{\left(-3\right)^3-1}=...\)
3.
\(\Leftrightarrow a+b=cot\left(\dfrac{A+B}{2}\right)\left(a.tanA+b.tanB\right)\)
\(\Leftrightarrow a\left(1-tanA.cot\left(\dfrac{A+B}{2}\right)\right)=b\left(tanB.cot\left(\dfrac{A+B}{2}\right)-1\right)\)
\(\Leftrightarrow a.\dfrac{cosA.sin\left(\dfrac{A+B}{2}\right)-sinA.cos\left(\dfrac{A+B}{2}\right)}{cosA.sin\left(\dfrac{A+B}{2}\right)}=b.\dfrac{sinB.cos\left(\dfrac{A+B}{2}\right)-cosB.sin\left(\dfrac{A+B}{2}\right)}{cosB.sin\left(\dfrac{A+B}{2}\right)}\)
\(\Leftrightarrow a.\dfrac{sin\left(\dfrac{B-A}{2}\right)}{cosA.sin\left(\dfrac{A+B}{2}\right)}=b.\dfrac{sin\left(\dfrac{B-A}{2}\right)}{cosB.sin\left(\dfrac{A+B}{2}\right)}\)
\(\Rightarrow\dfrac{a}{cosA}=\dfrac{b}{cosB}\Rightarrow\dfrac{2R.sinA}{cosA}=\dfrac{2R.sinB}{cosB}\)
\(\Rightarrow tanA=tanB\Rightarrow A=B\)
Tam giác ABC cân tại C
