1. If \(\angle A + \angle B = 90^\circ\), then prove that \[ 1 + \cfrac{\tan A}{\tan B} = \tan^2 A \sec^2 B \]

2. Given: \(\tan \theta + \sin \theta = m\) and \(\tan \theta - \sin \theta = n\) Prove that: \[ m^2 - n^2 = 4\sqrt{mn} \quad \text{where } 0^\circ < \theta < 90^\circ \]

3. If \( \tan 9^\circ = \frac{a}{b} \), then prove that \[ \frac{\sec^2 81^\circ}{1 + \cot^2 81^\circ} = \frac{b^2}{a^2} \]

4. If \(α\) and \(β\) are two complementary angles, show that \[ \sqrt{\frac{\tan α}{\cot β} + \tan α \cdot \cot β} = \sec α \]

5. Given \( \sin\alpha = m\sin\beta \) and \( \tan\alpha = n\tan\beta \), prove that \[ \sin\alpha = \sqrt{\frac{n^2 - m^2}{n^2 - 1}} \]

6. If \(∠P + ∠Q = 90^\circ\), then prove that \[ \sqrt{\frac{\sin P}{\cos Q}} - \sin P \cos Q = \cos^2 P \]

7. If \(\angle A + \angle B = 90^\circ\), then prove that \(1 + \frac{\tan A}{\tan B} = \csc^2 B\).

8. Prove that \[ \frac{\tan 47^\circ + \cot 27^\circ}{\tan 43^\circ + \cot 63^\circ} = \tan 47^\circ \cdot \cot 27^\circ \]

9. If ∠A + ∠B = 90°, then prove that \[ 1 + \frac{\tan A}{\tan B} = \sec^2 A \]

10. ABCD is a rectangle. By joining A and C, prove that \[ \tan^2 \angle CAD + 1 = \frac{1}{\sin^2 \angle BAC} \]

11. If \(\cot θ = 2\), then find the values of \(\tan θ\) and \(\sec θ\), and show that: \[1 + \tan^2θ = \sec^2θ\]

12. Prove that \[ \frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta} \]

13. Prove that \[ \cfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \cfrac{1 + \sin\theta}{\cos\theta} \] If you'd like, I can help walk through the proof step-by-step as well. Ready when you are!

14. If \( \sin \alpha = \frac{5}{13} \), then show that \[ \tan \alpha + \sec \alpha = \frac{3}{2} \]

15. If \(\angle P + \angle Q = 90^\circ\), then **prove** that: \[ \sqrt{\frac{\sin P}{\cos Q} - \sin P \cos Q} = \cos P \]

16. If \(a, b, c, d\) are in continued geometric progression, prove that \[(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2\]

17. If \(\frac{a^2}{b + c} = \frac{b^2}{c + a} = \frac{c^2}{a + b} = 1\), then prove that \[\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} = 1\]

18. If \(\frac{x}{y} = \frac{a + 2}{a - 2}\), then prove that \[ \frac{x^2 - y^2}{x^2 + y^2} = \frac{4a}{a^2 + 4} \]

19. Show that \[ \frac{2 \tan^2 30^\circ}{1 - \tan^2 30^\circ} + \sec^2 45^\circ - \cot^2 45^\circ = \sec 60^\circ \]

20. If \( \cot\theta = \frac{x}{y} \), then prove that \[ \frac{x\cos\theta - y\sin\theta}{x\cos\theta + y\sin\theta} = \frac{x^2 - y^2}{x^2 + y^2} \]

21. If \[ \frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} \] then prove that each of these ratios is either \(\frac{1}{2}\) or \(-1\).

22. In triangle △ABC, ∠C = 90°, and if BC = \(m\) and AC = \(n\), then prove that: \[ m \sin A + n \sin B = \sqrt{m^2 + n^2} \]

23. If \(bcx = cay = abz\), then prove that \[ \frac{ax + by}{a^2 + b^2} = \frac{by + cz}{b^2 + c^2} = \frac{cz + ax}{c^2 + a^2} \]

24. If \(x = cy + bz\), \(y = az + cx\), and \(z = bx + ay\), then prove that \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} \]

25. Left-hand side (LHS): \[ 1 + \cfrac{\tan A}{\tan B} = 1 + \cfrac{\tan(90^\circ - B)}{\tan B} = 1 + \cfrac{\cot B}{\cot B} = 1 + \cot^2 B = \csc^2 B \] Right-hand side (RHS): \[ \tan^2 A \cdot \sec^2 B = \tan^2(90^\circ - B) \cdot \sec^2 B = \cot^2 B \cdot \sec^2 B = \cfrac{\cos^2 B}{\sin^2 B} \cdot \cfrac{1}{\cos^2 B} = \cfrac{1}{\sin^2 B} = \csc^2 B \] \(\therefore\) LHS = RHS (Proved)

26. If \(a : b = b : c\), then prove that \[ (a + b)^2 : (b + c)^2 = a : c \]

27. If \(a^2 = by + cz\), \(b^2 = cz + ax\), and \(c^2 = ax + by\), then prove that \[ \frac{x}{a + x} + \frac{y}{b + y} + \frac{z}{c + z} = 1 \]

28. In triangle \(\triangle ABC\), prove that: \[ \sin\left(\frac{A + B}{2}\right) + \cos\left(\frac{B + C}{2}\right) = \cos\left(\frac{C}{2}\right) + \sin\left(\frac{A}{2}\right) \]

29. If \(a(\tan\theta + \cot\theta) = 1\) and \(\sin\theta + \cos\theta = b\), then prove that \(2a = b^2 - 1\), where \(0^\circ < \theta < 90^\circ\).

30. If \(a, b, c\) are in continued geometric progression (i.e., in successive proportion), then prove that \[ \cfrac{1}{b^2} = \cfrac{1}{b^2 - a^2} + \cfrac{1}{b^2 - c^2} \]