1. Let's analyze the expression $\frac{\sin(b/2) \sin(90+a)}{\cos(a/2) \sin(b)}$. 2. Recall that $\sin(90^\circ + x) = \cos(x)$ (using degrees). So, $\sin(90 + a) = \cos(a)$. 3. Substitute this into the expression: $$\frac{\sin\left(\frac{b}{2}\right) \cos(a)}{\cos\left(\frac{a}{2}\right) \sin(b)}$$ 4. Use the double-angle identity for sine: $\sin(b) = 2\sin\left(\frac{b}{2}\right)\cos\left(\frac{b}{2}\right)$. 5. Replace $\sin(b)$ in the denominator: $$\frac{\sin\left(\frac{b}{2}\right) \cos(a)}{\cos\left(\frac{a}{2}\right) \cdot 2 \sin\left(\frac{b}{2}\right) \cos\left(\frac{b}{2}\right)}$$ 6. Cancel $\sin\left(\frac{b}{2}\right)$ from numerator and denominator: $$\frac{\cos(a)}{2 \cos\left(\frac{a}{2}\right) \cos\left(\frac{b}{2}\right)}$$ 7. Final simplified form is $$ \frac{\cos(a)}{2 \cos\left(\frac{a}{2}\right) \cos\left(\frac{b}{2}\right)} $$.