Expression Evaluation A57Ca9

1. **State the problem:** We are given two equations: $$p \sin \theta + q \cos \theta = a$$ $$p \cos \theta - q \sin \theta = b$$ and we need to find the value of the expression: $$\frac{p+a}{q+b} + \frac{q-b}{p-a}$$ 2. **Rewrite the given equations:** From the first equation: $$a = p \sin \theta + q \cos \theta$$ From the second equation: $$b = p \cos \theta - q \sin \theta$$ 3. **Substitute $a$ and $b$ into the expression:** $$\frac{p + (p \sin \theta + q \cos \theta)}{q + (p \cos \theta - q \sin \theta)} + \frac{q - (p \cos \theta - q \sin \theta)}{p - (p \sin \theta + q \cos \theta)}$$ 4. **Simplify numerator and denominator of each fraction:** First fraction numerator: $$p + p \sin \theta + q \cos \theta = p(1 + \sin \theta) + q \cos \theta$$ First fraction denominator: $$q + p \cos \theta - q \sin \theta = p \cos \theta + q(1 - \sin \theta)$$ Second fraction numerator: $$q - p \cos \theta + q \sin \theta = -p \cos \theta + q(1 + \sin \theta)$$ Second fraction denominator: $$p - p \sin \theta - q \cos \theta = p(1 - \sin \theta) - q \cos \theta$$ 5. **Rewrite the expression:** $$\frac{p(1 + \sin \theta) + q \cos \theta}{p \cos \theta + q(1 - \sin \theta)} + \frac{-p \cos \theta + q(1 + \sin \theta)}{p(1 - \sin \theta) - q \cos \theta}$$ 6. **Multiply numerator and denominator of each fraction by the conjugate of the denominator to simplify or try to combine:** Instead, let's try to combine the two fractions over a common denominator: Common denominator: $$[p \cos \theta + q(1 - \sin \theta)][p(1 - \sin \theta) - q \cos \theta]$$ Numerator: $$[p(1 + \sin \theta) + q \cos \theta][p(1 - \sin \theta) - q \cos \theta] + [-p \cos \theta + q(1 + \sin \theta)][p \cos \theta + q(1 - \sin \theta)]$$ 7. **Expand each product:** First product: $$p(1 + \sin \theta) \cdot p(1 - \sin \theta) - p(1 + \sin \theta) q \cos \theta + q \cos \theta p(1 - \sin \theta) - q \cos \theta q \cos \theta$$ $$= p^2 (1 - \sin^2 \theta) - p q \cos \theta (1 + \sin \theta) + p q \cos \theta (1 - \sin \theta) - q^2 \cos^2 \theta$$ Note that: $$- p q \cos \theta (1 + \sin \theta) + p q \cos \theta (1 - \sin \theta) = p q \cos \theta[-(1 + \sin \theta) + (1 - \sin \theta)] = p q \cos \theta (-1 - \sin \theta + 1 - \sin \theta) = -2 p q \cos \theta \sin \theta$$ So first product simplifies to: $$p^2 (1 - \sin^2 \theta) - 2 p q \cos \theta \sin \theta - q^2 \cos^2 \theta$$ Second product: $$(-p \cos \theta)(p \cos \theta) + (-p \cos \theta) q (1 - \sin \theta) + q (1 + \sin \theta) p \cos \theta + q (1 + \sin \theta) q (1 - \sin \theta)$$ $$= -p^2 \cos^2 \theta - p q \cos \theta (1 - \sin \theta) + p q \cos \theta (1 + \sin \theta) + q^2 (1 - \sin^2 \theta)$$ Note that: $$- p q \cos \theta (1 - \sin \theta) + p q \cos \theta (1 + \sin \theta) = p q \cos \theta (-1 + \sin \theta + 1 + \sin \theta) = 2 p q \cos \theta \sin \theta$$ So second product simplifies to: $$- p^2 \cos^2 \theta + 2 p q \cos \theta \sin \theta + q^2 (1 - \sin^2 \theta)$$ 8. **Add the two simplified products:** $$[p^2 (1 - \sin^2 \theta) - 2 p q \cos \theta \sin \theta - q^2 \cos^2 \theta] + [- p^2 \cos^2 \theta + 2 p q \cos \theta \sin \theta + q^2 (1 - \sin^2 \theta)]$$ The $- 2 p q \cos \theta \sin \theta$ and $+ 2 p q \cos \theta \sin \theta$ cancel out. Group terms: $$p^2 (1 - \sin^2 \theta) - p^2 \cos^2 \theta + q^2 (1 - \sin^2 \theta) - q^2 \cos^2 \theta$$ Recall that: $$1 - \sin^2 \theta = \cos^2 \theta$$ So: $$p^2 \cos^2 \theta - p^2 \cos^2 \theta + q^2 \cos^2 \theta - q^2 \cos^2 \theta = 0$$ 9. **Result:** The numerator is zero, so the entire expression equals zero. **Final answer:** $$\boxed{0}$$