Question

Let $a,b$ and $c$ be the sides opposite to angles $A,$ $B$ and $C$ in a triangle $ABC.$ If $\sin A(\sin A+\cos B-\sin B)+\cos A(\cos A+\sin B+\cos B)=1+\sin C$ and $a=4,b=3,$ then area (in sq. units) of the triangle $ABC$ is

Answer 1

$\sin A(\sin A+\cos B-\sin B)+\cos A(\cos A+\sin B+\cos B)=1+\sin C$
$\Rightarrow ({\sin }^{2}A+{\cos }^{2}A)+(\sin A\cos B+\cos A\sin B)+(\cos A\cos B-\sin A\sin B)=1+\sin C$
$\Rightarrow 1+\sin (A+B)+\cos (A+B)=1+\sin C$
$\Rightarrow 1+\sin C-\cos C=1+\sin C(\because A+B+C=\pi )$
$\Rightarrow \cos C=0$
$\therefore C=\frac{\pi }{2}\therefore \Delta =\frac{1}{2}ab\sin C=\frac{1}{2}\times 4\times 3=6$