1. **Problem:** Given $\sec\theta + \tan\theta = p$, find the value of $\sin\theta$ in terms of $p$. 2. **Formula and rules:** Recall the identity: $$\sec^2\theta - \tan^2\theta = 1$$ Also, $\sec\theta = \frac{1}{\cos\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$. 3. **Step 1:** Square both sides of the given equation: $$ (\sec\theta + \tan\theta)^2 = p^2 $$ Expanding the left side: $$ \sec^2\theta + 2\sec\theta\tan\theta + \tan^2\theta = p^2 $$ 4. **Step 2:** Use the identity $\sec^2\theta - \tan^2\theta = 1$ to rewrite $\sec^2\theta + \tan^2\theta$: $$ \sec^2\theta + \tan^2\theta = (\sec^2\theta - \tan^2\theta) + 2\tan^2\theta = 1 + 2\tan^2\theta $$ So the equation becomes: $$ 1 + 2\tan^2\theta + 2\sec\theta\tan\theta = p^2 $$ 5. **Step 3:** Express $\sec\theta$ and $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$: $$ \sec\theta = \frac{1}{\cos\theta}, \quad \tan\theta = \frac{\sin\theta}{\cos\theta} $$ 6. **Step 4:** Substitute into the equation: $$ 1 + 2\left(\frac{\sin^2\theta}{\cos^2\theta}\right) + 2\left(\frac{1}{\cos\theta}\right)\left(\frac{\sin\theta}{\cos\theta}\right) = p^2 $$ Simplify: $$ 1 + 2\frac{\sin^2\theta}{\cos^2\theta} + 2\frac{\sin\theta}{\cos^2\theta} = p^2 $$ 7. **Step 5:** Multiply both sides by $\cos^2\theta$ to clear denominators: $$ \cos^2\theta + 2\sin^2\theta + 2\sin\theta = p^2 \cos^2\theta $$ 8. **Step 6:** Use $\sin^2\theta + \cos^2\theta = 1$ to replace $\cos^2\theta$: $$ (1 - \sin^2\theta) + 2\sin^2\theta + 2\sin\theta = p^2 (1 - \sin^2\theta) $$ Simplify left side: $$ 1 + \sin^2\theta + 2\sin\theta = p^2 - p^2 \sin^2\theta $$ 9. **Step 7:** Bring all terms to one side: $$ 1 + \sin^2\theta + 2\sin\theta - p^2 + p^2 \sin^2\theta = 0 $$ Group terms: $$ (1 - p^2) + (1 + p^2) \sin^2\theta + 2\sin\theta = 0 $$ 10. **Step 8:** Let $x = \sin\theta$. The quadratic equation is: $$ (1 + p^2) x^2 + 2x + (1 - p^2) = 0 $$ 11. **Step 9:** Solve for $x$ using quadratic formula: $$ x = \frac{-2 \pm \sqrt{4 - 4(1 + p^2)(1 - p^2)}}{2(1 + p^2)} $$ Calculate discriminant: $$ 4 - 4(1 + p^2)(1 - p^2) = 4 - 4(1 - p^4) = 4 - 4 + 4p^4 = 4p^4 $$ 12. **Step 10:** Substitute back: $$ x = \frac{-2 \pm 2p^2}{2(1 + p^2)} = \frac{-1 \pm p^2}{1 + p^2} $$ 13. **Step 11:** Since $0^\circ < \theta < 90^\circ$, $\sin\theta > 0$, so choose positive root: $$ \sin\theta = \frac{p^2 - 1}{p^2 + 1} $$ **Final answer:** $$ \boxed{\sin\theta = \frac{p^2 - 1}{p^2 + 1}} $$