In triangle \(\triangle ABC\), prove that: \[ \sin\left(\frac{A + B}{2}\right) + \cos\left(\frac{B + C}{2}\right) = \cos\left(\frac{C}{2}\right) + \sin\left(\frac{A}{2}\right) \]

Q.In triangle \(\triangle ABC\), prove that: \[ \sin\left(\frac{A + B}{2}\right) + \cos\left(\frac{B + C}{2}\right) = \cos\left(\frac{C}{2}\right) + \sin\left(\frac{A}{2}\right) \]

In triangle \(\triangle ABC\), we know: \[ A + B + C = 180^\circ \Rightarrow \frac{A + B + C}{2} = \frac{180^\circ}{2} = 90^\circ \Rightarrow \frac{A + B}{2} + \frac{C}{2} = 90^\circ \Rightarrow \frac{A + B}{2} = 90^\circ - \frac{C}{2} \] Again, \[ \frac{A + B + C}{2} = \frac{180^\circ}{2} = 90^\circ \Rightarrow \frac{B + C}{2} + \frac{A}{2} = 90^\circ \Rightarrow \frac{B + C}{2} = 90^\circ - \frac{A}{2} \] Now, the left-hand side of the identity is: \[ \sin\left(\frac{A + B}{2}\right) + \cos\left(\frac{B + C}{2}\right) = \sin\left(90^\circ - \frac{C}{2}\right) + \cos\left(90^\circ - \frac{A}{2}\right) = \cos\left(\frac{C}{2}\right) + \sin\left(\frac{A}{2}\right) \] Which equals the right-hand side. ( proved)