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

2. Evaluate the value of: \[ \frac{\sin 43^\circ \cdot \sin 65^\circ \cdot \tan 25^\circ + \sin 65^\circ \cdot \cot 65^\circ \cdot \cos 47^\circ}{\cos 65^\circ \cdot \sqrt{1 - \cos^2 43^\circ}} \]

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

4. 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)

5. Show that, \[ \csc^2 22^\circ \cdot \cot^2 68^\circ = \sin^2 22^\circ + \sin^2 68^\circ + \cot^2 68^\circ \]

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

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

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