Given that $\alpha + \beta = 90^{\circ}$,show that $\sqrt{\cos \alpha \operatorname{cosec} \beta - \cos \alpha \sin \beta} = \sin \alpha$.

(N/A) Given $\alpha + \beta = 90^{\circ}$,we have $\beta = 90^{\circ} - \alpha$.
Substituting $\beta$ in the expression:
$\sqrt{\cos \alpha \operatorname{cosec} \beta - \cos \alpha \sin \beta} = \sqrt{\cos \alpha \operatorname{cosec}(90^{\circ} - \alpha) - \cos \alpha \sin(90^{\circ} - \alpha)}$
Using trigonometric identities $\operatorname{cosec}(90^{\circ} - \alpha) = \sec \alpha$ and $\sin(90^{\circ} - \alpha) = \cos \alpha$:
$= \sqrt{\cos \alpha \sec \alpha - \cos \alpha \cos \alpha}$
Since $\cos \alpha \sec \alpha = 1$ and $\cos \alpha \cos \alpha = \cos^{2} \alpha$:
$= \sqrt{1 - \cos^{2} \alpha}$
Using the identity $1 - \cos^{2} \alpha = \sin^{2} \alpha$:
$= \sqrt{\sin^{2} \alpha} = \sin \alpha$.
Hence,the expression is proved.