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

2. 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)

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

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

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

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

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

8. In triangle ABC, \(\angle BAC\) is a right angle. If CD is the median, then prove that \[ BC^2 = CD^2 + 3AD^2 \]

9. 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!