Question: prove that $\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}$
1. **State the problem:** Prove that $$\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}$$ for an angle $\theta$.
2. **Recall definitions and identities:**
- $\tan \theta = \frac{\sin \theta}{\cos \theta}$
- $\sec \theta = \frac{1}{\cos \theta}$
3. **Rewrite the left-hand side (LHS) using these definitions:**
$$\frac{\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta} - \frac{1}{\cos \theta} + 1}$$
4. **Find a common denominator $\cos \theta$ in numerator and denominator:**
$$\frac{\frac{\sin \theta + 1 - \cos \theta}{\cos \theta}}{\frac{\sin \theta - 1 + \cos \theta}{\cos \theta}}$$
5. **Divide the fractions by multiplying numerator by reciprocal of denominator:**
$$\frac{\sin \theta + 1 - \cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta - 1 + \cos \theta} = \frac{\sin \theta + 1 - \cos \theta}{\sin \theta - 1 + \cos \theta}$$
6. **Simplify numerator and denominator by rearranging terms:**
Numerator: $\sin \theta + 1 - \cos \theta$
Denominator: $\sin \theta - 1 + \cos \theta$
7. **Rewrite denominator as:**
$$\sin \theta - 1 + \cos \theta = (\sin \theta + \cos \theta) - 1$$
8. **Rewrite numerator as:**
$$\sin \theta + 1 - \cos \theta = (\sin \theta - \cos \theta) + 1$$
9. **Try to factor or simplify numerator and denominator further. Instead, multiply numerator and denominator by the conjugate of denominator to simplify:**
Multiply numerator and denominator by $$\sin \theta + 1 - \cos \theta$$ (the numerator itself) to get:
$$\frac{(\sin \theta + 1 - \cos \theta)^2}{(\sin \theta - 1 + \cos \theta)(\sin \theta + 1 - \cos \theta)}$$
10. **Calculate denominator product using difference of squares pattern:**
$$(a - b)(a + b) = a^2 - b^2$$
Here, let $a = \sin \theta$, $b = 1 - \cos \theta$:
$$ (\sin \theta - (1 - \cos \theta))(\sin \theta + (1 - \cos \theta)) = \sin^2 \theta - (1 - \cos \theta)^2 $$
11. **Expand denominator:**
$$\sin^2 \theta - (1 - 2\cos \theta + \cos^2 \theta) = \sin^2 \theta - 1 + 2\cos \theta - \cos^2 \theta$$
12. **Use Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to simplify:**
$$\sin^2 \theta - \cos^2 \theta = 1 - 2\cos^2 \theta$$ but better to substitute directly:
$$\sin^2 \theta - 1 + 2\cos \theta - \cos^2 \theta = (\sin^2 \theta + \cos^2 \theta) - 1 + 2\cos \theta - 2\cos^2 \theta = 1 - 1 + 2\cos \theta - 2\cos^2 \theta = 2\cos \theta - 2\cos^2 \theta$$
13. **Factor denominator:**
$$2\cos \theta (1 - \cos \theta)$$
14. **Expand numerator:**
$$(\sin \theta + 1 - \cos \theta)^2 = (\sin \theta - \cos \theta + 1)^2 = (\sin \theta - \cos \theta)^2 + 2(\sin \theta - \cos \theta) + 1$$
15. **Calculate $(\sin \theta - \cos \theta)^2$:**
$$\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 1 - 2\sin \theta \cos \theta$$
16. **So numerator becomes:**
$$1 - 2\sin \theta \cos \theta + 2\sin \theta - 2\cos \theta + 1 = 2 - 2\sin \theta \cos \theta + 2\sin \theta - 2\cos \theta$$
17. **Factor numerator by 2:**
$$2(1 - \sin \theta \cos \theta + \sin \theta - \cos \theta)$$
18. **Rewrite numerator and denominator:**
$$\frac{2(1 - \sin \theta \cos \theta + \sin \theta - \cos \theta)}{2\cos \theta (1 - \cos \theta)}$$
19. **Cancel 2:**
$$\frac{1 - \sin \theta \cos \theta + \sin \theta - \cos \theta}{\cos \theta (1 - \cos \theta)}$$
20. **Rewrite numerator grouping terms:**
$$1 + \sin \theta - \cos \theta - \sin \theta \cos \theta$$
21. **Factor numerator as:**
$$(1 + \sin \theta)(1 - \cos \theta)$$
22. **So expression becomes:**
$$\frac{(1 + \sin \theta)(1 - \cos \theta)}{\cos \theta (1 - \cos \theta)}$$
23. **Cancel $(1 - \cos \theta)$:**
$$\frac{1 + \sin \theta}{\cos \theta}$$
24. **This matches the right-hand side (RHS), so the identity is proven.**
**Final answer:**
$$\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}$$