Question

Prove that $a^2\sin 2B+b^2\sin 2A=4\Delta .$

Answer 1

L.H.S.$=a^2\sin 2B+b^2\sin 2A$

$=4R^2\left[{\sin }^{2}A\right(2\sin B\cos B)+{\sin }^{2}B(2\sin A\cos A\left)\right]$ [using sine rule]

$=8R^2\sin A\sin B(\sin A\cos B+\sin B\cos A)$

$=8R^2\sin A\sin B\sin (A+B)$

$=8R^2\sin A\sin B\sin C=4\Delta$ .............. using sine rule $\sin A.\sin B.\sin C=4R\Delta$