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)

Q.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)

\[ \frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \] Or, \[ \frac{(\sin \theta - \cos \theta) + (\sin \theta + \cos \theta)}{(\sin \theta - \cos \theta) - (\sin \theta + \cos \theta)} = \frac{(1 - \sqrt{3}) + (1 + \sqrt{3})}{(1 - \sqrt{3}) - (1 + \sqrt{3})} \] [Using addition and subtraction process] Then, \[ \frac{2\sin \theta}{-2\cos \theta} = \frac{2}{-2\sqrt{3}} \Rightarrow \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} = \tan 30^\circ \therefore \theta = 30^\circ \quad \text{(Answer)} \]