4. Prove that \( \frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=1+\tan A+\cot A=1+\sec A \operatorname{cosec} A \)

To prove the identity \[ \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \tan A + \cot A = 1 + \sec A \csc A, \] we will start by simplifying the left-hand side: 1. **Rewrite \(\tan A\) and \(\cot A\)**: \[ \tan A = \frac{\sin A}{\cos A}, \quad \cot A = \frac{\cos A}{\sin A}. \] 2. **Substituting into the left-hand side**: \[ \frac{\tan A}{1 - \cot A} = \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} = \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} = \frac{\sin^2 A}{\cos A (\sin A - \cos A)}. \] Similarly, \[ \frac{\cot A}{1 - \tan A} = \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}} = \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}} = \frac{\cos^2 A}{\sin A (\cos A - \sin A)}. \] 3. **Combine the two fractions**: \[ \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = \frac{\sin^2 A}{\cos A (\sin A - \cos A)} + \frac{\cos^2 A}{\sin A (\cos A - \sin A)}. \] Notice that \(\cos A - \sin A = -(\sin A - \cos A)\), so we can rewrite the second term: \[ \frac{\cos^2 A}{\sin A (\cos A - \sin A)} = -\frac{\cos^2 A}{\sin A (\sin A - \cos A)}. \] Thus, we have: \[ \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = \frac{\sin^2 A}{\cos A (\sin A - \cos A)} - \frac{\cos^2 A}{\sin A (\sin A - \cos A)}. \] Combine the fractions: \[ = \frac{\sin^2 A \sin A - \cos^2 A \cos A}{\sin A \cos A (\sin A - \cos A)} = \frac{\sin^3 A - \cos^3 A}{\sin A \cos A (\sin A - \cos A)}. \] 4. **Factor the numerator**: The numerator \(\sin^3 A - \cos^3 A\) can be factored using the difference of cubes: \[ \sin^3 A - \cos^3 A = (\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A). \] Since \(\sin^2 A + \cos^2 A = 1\), we have: \[ \sin^3 A - \cos^3 A = (\sin A - \cos A)(1 + \sin A \cos A). \] 5. **Substituting back**: Substitute this back into our expression: \[ \frac{(\sin A - \cos A)(1 + \sin A \cos A)}{\sin A \cos A (\sin A - \cos A)}. \] The \((\sin A - \cos A)\) terms cancel (assuming \(A \neq \frac{\pi}{4} + n\frac{\pi}{2}\)): \[ = \frac{1 + \sin A \cos A}{\sin A \cos A}. \] 6. **Simplifying further**: This can be rewritten as: \[ = \frac{1}{\sin A \cos A} + 1 = \csc A \sec A + 1. \] 7. **Final result**: Thus, we have shown that: \[ \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1