1. **State the problem:** Given $\sin x = \frac{p - q}{p + q}$ where $x$ is between $0^\circ$ and $90^\circ$, find $1 - \tan^2 x$. 2. **Recall the identity:** We know that $1 - \tan^2 x = \frac{\cos 2x}{\cos^2 x}$ or more simply, using the Pythagorean identity, $1 - \tan^2 x = \frac{\cos 2x}{\cos^2 x}$, but a more straightforward approach is to use the identity $1 - \tan^2 x = \frac{1 - \sin^2 x}{\cos^2 x} - \tan^2 x$ which simplifies to $\cos 2x$. 3. **Use the identity:** Actually, the simplest identity is: $$1 - \tan^2 x = \frac{\cos 2x}{\cos^2 x}$$ But since $x$ is in the first quadrant, $\cos x > 0$, so we can use the double angle formula: $$\cos 2x = 1 - 2\sin^2 x$$ 4. **Calculate $\sin x$:** Given $\sin x = \frac{p - q}{p + q}$. 5. **Calculate $\sin^2 x$:** $$\sin^2 x = \left(\frac{p - q}{p + q}\right)^2 = \frac{(p - q)^2}{(p + q)^2}$$ 6. **Calculate $\cos 2x$:** $$\cos 2x = 1 - 2\sin^2 x = 1 - 2 \times \frac{(p - q)^2}{(p + q)^2} = \frac{(p + q)^2 - 2(p - q)^2}{(p + q)^2}$$ 7. **Expand numerator:** $$(p + q)^2 = p^2 + 2pq + q^2$$ $$(p - q)^2 = p^2 - 2pq + q^2$$ 8. **Substitute:** $$\cos 2x = \frac{p^2 + 2pq + q^2 - 2(p^2 - 2pq + q^2)}{(p + q)^2} = \frac{p^2 + 2pq + q^2 - 2p^2 + 4pq - 2q^2}{(p + q)^2}$$ 9. **Simplify numerator:** $$= \frac{-p^2 + 6pq - q^2}{(p + q)^2}$$ 10. **Recall that $1 - \tan^2 x = \cos 2x$ for $x$ in $0^\circ$ to $90^\circ$.** **Final answer:** $$1 - \tan^2 x = \frac{-p^2 + 6pq - q^2}{(p + q)^2}$$