### If ^@\alpha + \beta = 90^ \circ^@, simplify ^@ \sqrt{\cos \alpha \mathrm{cosec} \beta - \cos \alpha \sin \beta}. ^@

**Answer:**

^@ \pm sin \alpha ^@

**Step by Step Explanation:**

- It is given that ^@ \alpha + \beta = 90^ \circ ^@

^@ \implies \beta = 90^ \circ - \alpha ^@ - Replace ^@\beta ^@ with ^@90^ \circ - \alpha^@ in given expression.

^@ \begin{align} & \sqrt{\cos \alpha \mathrm{cosec} \beta - \cos \alpha \sin \beta} \\ = & \sqrt{\cos \alpha\mathrm{cosec}(90^{\circ}-\alpha) - \cos \alpha \sin(90^{\circ}-\alpha)} \\ = & \sqrt{\cos \alpha\sec \alpha - \cos \alpha \cos \alpha} \\ = & \sqrt{1 - \cos^2 \alpha} \\ = & \pm sin \alpha \end{align}^@