For the triangle ABC, show that

$\text{sin}^2\dfrac{A}{2}$ + $\text{sin}^2\dfrac{B+C}{2}$ = 1.

For triangle ABC,

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ B + ∠ C = 180° - ∠ A

⇒ $\dfrac{B + C}{2} = \dfrac{180° - A}{2}$

⇒ $\dfrac{B + C}{2} = 90° - \dfrac{A}{2}$

$\text{L.H.S.} = \text{sin}^2\dfrac{A}{2} + \text{sin}^2\dfrac{B+C}{2}\\[1em] = \text{sin}^2\dfrac{A}{2} + \text{sin}^2\Big(90° - \dfrac{A}{2}\Big)\\[1em] = \text{sin}^2\dfrac{A}{2} + \text{cos}^2\dfrac{A}{2}\\[1em] = 1$

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence, $\text{sin}^2\dfrac{A}{2}$ + $\text{sin}^2\dfrac{B+C}{2}$ = 1.