In a triangle $ABC$, if $3\sin A+4\cos B=6$ and $4\sin B+3\cos A=1$, then $\sin (A+B)$ is equal to

Given, $3\sin A+4\cos B=6$ and $4\sin B+3\cos A=1$

On squaring and adding both the equations we get,

${\left(3\sin A+4\cos B\right)}^2+{\left(4\sin B+3\cos A\right)}^2=6^2+1^2$

$\Rightarrow 9{\sin }^{2}A+16{\cos }^{2}B+24\sin A\cos B+16{\sin }^{2}B+9{\cos }^{2}A+24\sin B\cos A=37$

$\Rightarrow 9\left({\sin }^{2}A+{\cos }^{2}A\right)+16\left({\cos }^{2}B+{\sin }^{2}B\right)+24\left(\sin A\cos B+\sin B\cos A\right)=37$

Using ${\sin }^{2}A+{\cos }^{2}A=1$ and $\sin (A+B)=\sin A\cos B+\cos A\sin B$

$\Rightarrow 9+16+24\sin \left(A+B\right)=37$

$\Rightarrow 24\sin \left(A+B\right)=37-25=12$

$\Rightarrow \sin (A+B)=\frac{1}{2}$