1. Prove that: \( \csc^2 22^\circ \cdot \cot^2 68^\circ = \sin^2 22^\circ + \sin^2 68^\circ + \cot^2 68^\circ \)

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

3. Prove: \[ \csc^2 22^\circ \cdot \cot^2 68^\circ = \sin^2 22^\circ + \sin^2 68^\circ + \cot^2 68^\circ \]

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)