Mỏi cổ quá, em cố gắng xoay đề lại cho mọi người dễ đọc nha
\(\dfrac{sin^3\dfrac{B}{2}}{cos\left(\dfrac{A+C}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(\dfrac{A+C}{2}\right)}-\dfrac{cos\left(A+C\right)}{sinB}.tanB\)
\(=\dfrac{sin^3\dfrac{B}{2}}{cos\left(\dfrac{180^0-B}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(\dfrac{180^0-B}{2}\right)}-\dfrac{cos\left(180^0-B\right)}{sinB}.tanB\)
\(=\dfrac{sin^3\dfrac{B}{2}}{cos\left(90^0-\dfrac{B}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(90^0-\dfrac{B}{2}\right)}+\dfrac{cosB}{sinB}.tanB\)
\(=\dfrac{sin^3\dfrac{B}{2}}{sin\left(\dfrac{B}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{cos\dfrac{B}{2}}+cotB.tanB\)
\(=sin^2\dfrac{B}{2}+cos^2\dfrac{B}{2}+1=1+1=2\)