Explanation

(b)
We have, $\sin^{- 1}{x + \sin^{- 1}{y = \pi - \sin^{- 1}\mathcal{z}}}$
or $x\sqrt{1 - y^{2}} + y\sqrt{1 - x^{2}\mathcal{= z}}$
or $x^{2}\left( 1 - y^{2} \right) = \mathcal{z}^{2} + y^{2}\left( 1 - x^{2} \right) - 2y\mathcal{z}\sqrt{(1 - x^{2})}$
or $\left( x^{2} - \mathcal{z}^{2} - y^{2} \right)^{2} = 4y^{2}\mathcal{z}^{2}(1 - x^{2})$
or $x^{4} + y^{4} + \mathcal{z}^{4} - 2x^{2}\mathcal{z}^{2} + 2y^{2}\mathcal{z}^{2} - 2x^{2}y^{2} + 4x^{2}y^{2}\mathcal{z}^{2} - 4y^{2}\mathcal{z}^{2} = 0$
or $x^{4} + y^{4} + \mathcal{z}^{4} + 4x^{2}y^{2}\mathcal{z}^{2} = 2(x^{2}y^{2} + y^{2}\mathcal{z}^{2} + \mathcal{z}^{2}x^{2})$
$\therefore\ k = 2$