State whether the following statement is true or false.

In $△ABC$, $\sin \left(\frac{A+B}{2}\right)=\cos \frac{C}{2}$.

The correct option is A True
In any triangle, $ABC$, the sum of angles will be ${180}^0=\pi$.

Hence,

$A+B+C={180}^0$

$⟹A+B=\pi -C$

$⟹\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}$

Now,

$LHS=sin\left(\frac{A+B}{2}\right)=sin(\frac{\pi }{2}-\frac{C}{2})=cos\frac{C}{2}=RHS.$.......................[Since, $sin(\pi /2-\theta )=cos\theta$]